Chapter 7

A Complete Resource Book in Chemistry for JEE Main · 157 exercises

Problem 125

A. 1 M solution of glucose reaches dissociation equilibrium given below. $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \rightleftharpoons 6 \mathrm{HCHO} $$ If the equilibrium constant is \(0.167 \times 10^{-22}\), the concentration of \(\mathrm{HCHO}\) in the equilibrium is (a) \(1.60 \times 10^{-8} \mathrm{M}\) (b) \(3.20 \times 10^{-6} \mathrm{M}\) (c) \(3.20 \times 10^{-4} \mathrm{M}\) (d) \(1.60 \times 10^{-4} \mathrm{M}\)

5 step solution

Problem 126

For the dissociation reaction \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g})\) the degree of dissociation \((\alpha)\) in terms of \(K_{p}\) and total equilibrium pressure \(\mathrm{P}\) is: (a) \(\alpha=\sqrt{\frac{4 \mathrm{P}+\mathrm{K}_{\mathrm{p}}}{\mathrm{K}_{\mathrm{P}}}}\) (b) \(\alpha=\sqrt{\frac{\mathrm{K}_{\mathrm{p}}}{4 \mathrm{P}+\mathrm{K}_{\mathrm{p}}}}\) (c) \(\alpha=\sqrt{\frac{\mathrm{K}_{\mathrm{p}}}{4 \mathrm{P}}}\) (d) None of these

5 step solution

Problem 128

When \(\mathrm{CO}_{2}\) dissolves in water, the following equilibrium is established. \(\mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{CO}_{3}^{-}\) The equilibrium constant is \(3.8 \times 10^{-7}\) and at \(\mathrm{pH}=6.0\), the ratio \(\frac{\left[\mathrm{HCO}_{3}^{-}\right]}{\left[\mathrm{CO}_{2}\right]}\) will be (a) \(3.8 \times 10^{-13}\) (b) \(3.8 \times 10^{-1}\) (c) \(6.0\) (d) \(13.4\)

6 step solution

Problem 129

At certain temperature compound \(\mathrm{AB}_{2}(\mathrm{~g})\) dissociates according to the reaction \(2 \mathrm{AB}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{AB}(\mathrm{g})+\mathrm{B}_{2}(\mathrm{~g})\) The value of \(K_{p}\) in terms of degree of dissociation ' \(\alpha\) ' and total pressure ' \(\mathrm{P}\) ' is (a) \(\mathrm{P} \frac{\alpha^{3}}{2}\) (b) \(\mathrm{P} \frac{\alpha^{2}}{3}\) (c) \(\mathrm{P} \frac{\alpha^{3}}{3}\) (d) \(\mathrm{P} \frac{\alpha^{2}}{2}\)

7 step solution

Problem 130

When \(\mathrm{NaNO}_{3}(\mathrm{~d}=2.0 \mathrm{~g} / \mathrm{cc})\) is heated in a closed vessel of \(100 \mathrm{ml}\), oxygen is liberated and \(\mathrm{NaNO}_{2}\) \((\mathrm{d}=1.5 \mathrm{~g} / \mathrm{cc})\) is left behind as per the reaction : \(2 \mathrm{NaNO}_{3}(\mathrm{~s}) \rightleftharpoons 2 \mathrm{NaNO}_{2}(\mathrm{~s})+\mathrm{O}_{2}(\mathrm{~g})\). At equilibrium, the volumes of \(\mathrm{NaNO}_{3}\) left and \(\mathrm{NaNO}_{3}\) left and \(\mathrm{NaNO}_{2}\) produced are very small and can be neglected. Which of the following is a correct statement about this equilibrium? (a) Addition of \(30 \mathrm{~g}\) of \(\mathrm{NaNO}_{3}\) favours reverse reaction. (b) Addition of \(30 \mathrm{~g}\) of \(\mathrm{NaNO}_{2}\) favours forward reaction. (c) Increasing temperature favours reverse reaction. (d) None of these.

6 step solution

Problem 131

An element \(\mathrm{X}\) being with oxygen and CO present in air as \(\mathrm{X}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{XO}_{2}(\mathrm{~g}) ; \mathrm{K}_{p_{1}}=27\) \(\mathrm{X}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{XCO}(\mathrm{g}) ; \quad \mathrm{K}_{\mathrm{P}_{2}}=10^{4}\) When \(\mathrm{X}\) at 1 atm is treated with air, \(25 \%\) of it is bound to \(\mathrm{CO}(\mathrm{g})\). The partial pressure of \(\mathrm{CO}(\mathrm{g})\) in air at equilibrium, if partial pressure of \(\mathrm{O}_{2}(\mathrm{~g})\) in air at equilibrium is \(0.2 \mathrm{~atm}\), would be (a) \(1.9 \times 10^{4} \mathrm{~atm}\) (b) \(1.9 \times 10^{-4} \mathrm{~atm}\) (c) \(2.08 \times 10^{4} \mathrm{~atm}\) (d) \(2.08 \times 10^{-4} \mathrm{~atm}\)

4 step solution

Problem 132

For the reaction (1) and (2) \(\mathrm{A} \rightleftharpoons(\mathrm{g}) \quad \mathrm{B}(\mathrm{g})+\mathrm{C}(\mathrm{g})\) \(\mathrm{X} \rightleftharpoons(\mathrm{g}) \quad 2 \mathrm{Y}(\mathrm{g})\) Given, \(K_{p_{1}}: K_{p_{2}}=9: 1 .\) If degree of dissociation of \(A\) (g) and \(\mathrm{X}\) (g) are same then the ratio of total pressure in equilibrium (1) and (2) will be (a) \(36: 1\) (b) \(0.5: 1\) (c) \(1: 1\) (d) \(3: 1\)

5 step solution

Problem 134

For the reaction \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\) at \(500^{\circ} \mathrm{C}\), the value of \(\mathrm{K}_{\mathrm{p}}\) is \(1.44 \times 10^{-5}\) when the partial pressures are measured in atmosphere. The value of \(\mathrm{K}_{\mathrm{c}}\) with concentration in \(\mathrm{mol} \mathrm{L}^{-1}\) is (a) \(\frac{1.44 \times 10^{-5}}{(8.314 \times 773)^{-2}}\) (b) \(\frac{1.44 \times 10^{-5}}{(0.082 \times 500)^{-2}}\) (c) \(\frac{1.44 \times 10^{-5}}{(0.082 \times 773)^{-2}}\) (d) \(\frac{1.44 \times 10^{-5}}{(0.082 \times 773)^{-2}}\)

7 step solution

Problem 135

The degree of dissociation of \(\mathrm{PCl}_{5}\) at one atmosphere is \(0.3 .\) The pressure at which \(\mathrm{PCl}_{5}\) is dissociated to \(50 \%\) is (a) \(0.273 \mathrm{~atm}\) (b) \(0.3 \mathrm{~atm}\) (c) \(1.34 \mathrm{~atm}\) (d) \(2.73 \mathrm{~atm}\)

7 step solution

Problem 136

At \(550 \mathrm{~K}\), the \(\mathrm{Kc}\) for the following reaction is \(10^{4}\) \(\mathrm{mol}^{-1} \mathrm{~L}\) \(\mathrm{X}(\mathrm{g})+\mathrm{Y}(\mathrm{g}) \rightleftharpoons \mathrm{Z}(\mathrm{g})\) At equilibrium, it was observed that \([\mathrm{X}]=\frac{1}{2}[\mathrm{Y}]=\frac{1}{2}[\mathrm{Z}]\) What is the value of \([\mathrm{Z}]\) at equilibrium? (a) \(10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}\) (b) \(2 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}\) (c) \(2 \times 10^{4} \mathrm{~mol} \mathrm{~L}^{-1}\) (d) \(10^{4} \mathrm{~mol} \mathrm{~L}^{-1}\)

4 step solution

Problem 137

For a chemical reaction \(\mathrm{A} \longrightarrow\) product, the mechanism of the reaction postulated was as follows. $$ \mathrm{A} \stackrel{\mathrm{k}_{1}}{\mathrm{~g}_{2}} 3 \mathrm{~B} \frac{\mathrm{k}_{\mathrm{s}}}{\text { R.D. }}{\mathrm{\longrightarrow}} \mathrm{C}_{\mathrm{g}} $$ If the reaction occurred with individual rate constants \(\mathrm{k}_{1}, \mathrm{k}_{2}\) and \(\mathrm{k}_{3}\), determine activation energy for the overall reaction if the activation energies associated with these rate constants are \(\mathrm{E}_{a_{1}}=180 \mathrm{~kJ} \mathrm{~mol}^{-1}, \mathrm{E}_{a_{2}}=90 \mathrm{~kJ}\) \(\mathrm{mol}^{-1}\) and \(\mathrm{E}_{a_{3}}=40 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (a) \(70 \mathrm{~kJ}\) (b) \(-10 \mathrm{~kJ}\) (c) \(310 \mathrm{~kJ}\) (d) \(130 \mathrm{~kJ}\)

5 step solution

Problem 139

For the reversible system \(\mathrm{PCl}_{3}+\mathrm{Cl}_{2} \rightleftharpoons \mathrm{PCl}_{5}\), the relationship between degree of dissociation \((\alpha)\) of \(\mathrm{PCl}_{5}\) and the equilibrium constant \(K_{p}\) of the above equilibrium is (a) \(\frac{\mathrm{K}_{\mathrm{p}}}{\sqrt{\mathrm{P}+\mathrm{K}_{\mathrm{p}}}}\) (b) \(\sqrt{\frac{\mathrm{K}_{\mathrm{p}}}{1+\mathrm{P} \cdot \mathrm{K}_{\mathrm{p}}}}\) (c) \(\frac{1}{\sqrt{\mathrm{P}+\mathrm{K}_{\mathrm{P}}}}\) (d) \(\sqrt{\frac{1}{1+P \cdot K_{P}}}\)

5 step solution

Problem 140

One mole of \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})\) at \(300 \mathrm{~K}\) is kept in a closed container under one atmospheric pressure. It is heated to \(600 \mathrm{~K}\) when \(20 \%\) by mass of \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})\) decomposes to \(\mathrm{NO}_{2}(\mathrm{~g})\). The resultant pressure is (a) \(1.0 \mathrm{~atm}\) (b) \(2.4 \mathrm{~atm}\) (c) \(2.0 \mathrm{~atm}\) (d) \(1.2 \mathrm{~atm}\)

7 step solution

Problem 141

The equilibrium constant (K) of the reaction, \(\mathrm{A}+\mathrm{B}=\mathrm{C}+\mathrm{D}\) at \(298 \mathrm{~K}\) is \(100 .\) If the rate con- stant of the forward reaction is \(4 \times 10^{5}\), the rate constant of the reverse reaction is (a) 4 (b) \(4 \times 10^{2}\) (c) \(4 \times 10^{3}\) (d) \(4 \times 10^{5}\)

5 step solution

Problem 142

One mole of \(\mathrm{N}_{2} \mathrm{O}_{4}\) (g) at \(310 \mathrm{~K}\) is \(25 \%\) dissociated at 1 atm pressure. The percentage dissociation at \(0.1 \mathrm{~atm}\) and \(310 \mathrm{~K}\) is (a) \(25 \%\) (b) \(50 \%\) (c) \(76 \%\) (d) \(63 \%\)

5 step solution

Problem 144

\(K_{p}\) has the value of \(10^{-6} \mathrm{~atm}^{3}\) and \(10^{-4} \mathrm{~atm}^{3}\) at \(298 \mathrm{~K}\) and \(323 \mathrm{~K}\) respectively for the reaction \(\mathrm{CuSO}_{4} \cdot 3 \mathrm{H}_{2} \mathrm{O}(\mathrm{s}) \rightleftharpoons \mathrm{CuSO}_{4}(\mathrm{~s})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) \(\Delta_{r} H^{\circ}\) for the reaction is: (a) \(7.7 \mathrm{~kJ} / \mathrm{mol}\) (b) \(-147.41 \mathrm{~kJ} / \mathrm{mol}\) (c) \(147.41 \mathrm{~kJ} / \mathrm{mol}\) (d) None of these

5 step solution

Problem 145

\(\mathrm{K}_{\mathrm{p}}\) has the value of \(10^{-6} \mathrm{~atm}^{3}\) and \(10^{-4} \mathrm{~atm}^{3}\) at \(298 \mathrm{~K}\) and \(323 \mathrm{~K}\) respectively for the reaction \(\mathrm{CuSO}_{4} \cdot 3 \mathrm{H}_{2} \mathrm{O}(\mathrm{s}) \rightleftharpoons \mathrm{CuSO}_{4}(\mathrm{~s})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) \(\Delta_{r} \mathrm{H}^{\circ}\) for the reaction is: (a) \(7.7 \mathrm{~kJ} / \mathrm{mol}\) (b) \(-147.41 \mathrm{~kJ} / \mathrm{mol}\) (c) \(147.41 \mathrm{~kJ} / \mathrm{mol}\) (d) None of these

6 step solution

Problem 146

In a system \(\mathrm{A}(\mathrm{s}) \rightleftharpoons 2 \mathrm{~B}(\mathrm{~g})+3 \mathrm{C}(\mathrm{g})\) If the concentration of \(\mathrm{C}\) at equilibrium is increased by a factor of 2 , it will cause the equilibrium concentration of \(\mathrm{B}\) to change to (a) One half of its original value (b) Two times the original value (c) \(\frac{1}{2 \sqrt{2}}\) times the original value (d) \(2 \sqrt{2}\) times its original value

6 step solution

Problem 147

\(\frac{\mathrm{K}_{\mathrm{p}}}{\mathrm{K}_{\mathrm{C}}}\) for the reaction \(\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{~g})\) will be (a) \(\sqrt{\mathrm{RT}}\) (b) \(\mathrm{RT}\) (c) \(\frac{1}{\sqrt{\mathrm{RT}}}\) (d) 1

4 step solution

Problem 148

At \(25^{\circ} \mathrm{C}\), the standard emf of a cell having reaction involving two electron exchange is found to be \(0.295 \mathrm{~V}\). The equilibrium constant of the reaction is approximately (a) \(9.50 \times 10^{9}\) (b) \(1 \times 10^{10}\) (c) 10 (d) \(9.51 \times 10^{7}\)

6 step solution

Problem 149

Assume that the decomposition of \(\mathrm{HNO}_{3}\) can be represented by the following equation \(4 \mathrm{HNO}_{3}(\mathrm{~g}) \rightleftharpoons 4 \mathrm{NO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g})\) and the reaction approaches equilibrium at \(400 \mathrm{~K}\) temperature and \(30 \mathrm{~atm}\) pressure. At equilibrium, partial pressure of \(\mathrm{HNO}_{3}\) is \(2 \mathrm{~atm} .\) Calculate \(\mathrm{K}_{c}\) in \((\mathrm{mol} / \mathrm{L})^{3}\) at \(400 \mathrm{~K}\) : (Use : \(\mathrm{R}=0.08 \mathrm{~atm}-\mathrm{L} / \mathrm{mol}-\mathrm{K})\) (a) 4 (b) 8 (c) 16 (d) 32

6 step solution

Problem 151

The equilibrium constants for the reactions \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\) and \(\frac{1}{2} \mathrm{~N}_{2}+\frac{3}{2} \mathrm{H}_{2} \rightleftharpoons \mathrm{NH}_{3}\) are \(K_{1}\) and \(K_{2}\) respectively. The correct relationship between \(\mathrm{K}_{1}\) and \(\mathrm{K}_{2}\) is (a) \(\mathrm{K}_{\mathrm{t}}=\frac{\mathrm{K}_{2}}{2}\) (b) \(\mathrm{K}_{2}=\sqrt{\mathrm{K}_{1}}\) (c) \(\mathrm{K}_{2}=\mathrm{K}_{1}{\underline{\phantom{xx}}}^{2}\) (d) \(\mathrm{K}_{1}=\mathrm{K}_{2}\)

4 step solution

Problem 152

The coagulation of \(100 \mathrm{ml}\) of colloidal solution of gold is completely prevented by addition of \(0.25 \mathrm{~g}\) of a substance " \(X\) " to it before addition of \(10 \mathrm{ml}\) of \(10 \% \mathrm{NaCl}\) solution. The gold number of "' \(\mathrm{X}\) " is (a) 25 (b) 250 (c) \(2.5\) (d) \(0.25\)

5 step solution

Problem 153

One mole of \(\mathrm{N}_{2}(\mathrm{~g})\) is mixed with 2 moles of \(\mathrm{H}_{2}(\mathrm{~g})\) in a 4 litre vessel. If \(50 \%\) of \(\mathrm{N}_{2}(\mathrm{~g})\) is converted to \(\mathrm{NH}_{3}(\mathrm{~g})\) by the following reaction: $$ \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) $$ What will be the value of \(K_{c}\) for the following equilibrium? $$ \mathrm{NH}_{3}(\mathrm{~g}) \rightleftharpoons \frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g})+\frac{3}{2} \mathrm{H}_{2}(\mathrm{~g}) $$ (a) 256 (b) 16 (c) \(\frac{1}{16}\) (d) None of these

6 step solution

Problem 156

Which of the following will favour the formation of \(\mathrm{NH}_{3}\) by Haber's Process? (a) Increase of temperature (b) Increase of pressure (c) Addition of catalyst (d) Addition of promoter

5 step solution

Problem 157

When solid \(\mathrm{NaNO}_{3}\) is heated in a closed vessel, \(\mathrm{O}_{2}\) is liberated and solid \(\mathrm{NaNO}_{2}\) is left behind. At equilibrium (a) Addition of \(\mathrm{NaNO}_{2}\) favours reverse reaction. (b) Addition of \(\mathrm{NaNO}_{3}\) favours forward reaction. (c) Increasing the pressure favours reverse reaction. (d) Increasing the temperature favours forward reaction.

5 step solution

Problem 158

\(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\) At constant temperature, forward reaction is favoured by (a) introducing inert gas at constant volume(b) introducing inert gas at constant pressure (c) introducing chlorine gas at constant volume (d) increasing the volume of the container

5 step solution

Problem 160

If the concentrations of two monobasic acids are same, their relative strengths can be compared by (a) \(\left(\frac{\mathrm{K}_{1}}{\mathrm{~K}_{2}}\right)\) (b) \(\left(\frac{\alpha_{1}}{\alpha_{2}}\right)\) (c) \(\left(\sqrt{\frac{K_{1}}{K_{2}}}\right)\) (d) \(\frac{\left[\mathrm{H}^{+}\right]_{1}}{\left[\mathrm{H}^{+}\right]_{2}}\)

5 step solution

Problem 167

Match the following Column-I (a) \(\mathrm{N}_{2} \mathrm{O}_{4(\mathrm{~g})} \rightleftharpoons 2 \mathrm{NO}_{2(g)}\) (b) \(\mathrm{PCl}_{s(\mathrm{~g})} \rightleftharpoons \mathrm{PCl}_{3(\mathrm{~g})}+\mathrm{Cl}_{2(\mathrm{~g})}\) (c) \(\mathrm{NH}_{2} \mathrm{COONH}_{4(\mathrm{~s})} \rightleftharpoons 2 \mathrm{NH}_{3(\mathrm{~g})}+\mathrm{CO}_{2 \mathrm{~g}}\) (d) \(\mathrm{H}_{2(g)}+\mathrm{I}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{HI}_{(\mathrm{g})}\) Column-II (p) \(K_{P}=\frac{4 x^{2} p}{a^{2}-x^{2}}\) (q) \(K_{P}=\frac{4 P^{3}}{27}\) (r) \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}\) (s) \(K_{p}=K_{c^{\circ}} R T\) (t) \(\Delta \mathrm{n}_{\mathrm{g}}>0\)

4 step solution

Problem 169

Match the following Column-I (a) \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\) (b) \(\mathrm{CaCO}_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{CaO}\) (s) \(+\mathrm{CO}_{2}(\mathrm{~g})\) (c) \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g})=2 \mathrm{NH}_{3}(\mathrm{~g})\) (d) \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\) Column-II (p) Unaffected by inert gas addition at constant volume (q) Forward shift by rise in pressure (r) Unaffected by increase in pressure (s) Backward shift by rise in pressure (t) reaction has \(\Delta n_{g}>0\)

4 step solution

Problem 171

Assertion: The equilibrium of \(\mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{B}(\mathrm{g})+\mathrm{C}(\mathrm{g})\) is not affected by changing volume of the system. Reason: \(\mathrm{K}_{\mathrm{c}}\) for the reaction does not depend on volume of the container.

4 step solution

Problem 172

Assertion: Adding an inert gas todissociation equilibrium of gaseous \(\mathrm{N}_{2} \mathrm{O}_{4}\) at constant pressure and temperature increases the dissociation. Reason: Molar concentration of the reactants and products decreases on the addition of inert gas.

6 step solution

Problem 176

The following mechanism proposed for a reaction \(2 \mathrm{~A}+\mathrm{B} \rightarrow \mathrm{D}+\mathrm{E}\) is as \(\mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{C}\) (fast) \(\mathrm{A}+\mathrm{C} \rightarrow \mathrm{E}+\mathrm{D}(\mathrm{slow})\) The order of reaction is

5 step solution

Problem 182

In the heterogeneous equilibrium \(\mathrm{A}(\mathrm{s}) \rightleftharpoons 2 \mathrm{~B}(\mathrm{~g})+\) \(3 \mathrm{C}(\mathrm{g})\). If conc. of \(\mathrm{C}\) at equilibrium is doubled, then after the equilibrium is re- established, the conc. of \(\mathrm{B}\) will be \(\left(\frac{1}{\mathrm{r}}\right)\) times its original value. The value of \(\mathrm{r}^{2}\) is

5 step solution

Problem 184

For the equilibrium \(\mathrm{AB}(\mathrm{g}) \rightleftharpoons \mathrm{A}(\mathrm{g})+\mathrm{B}(\mathrm{g})\), at a given temperature \(\frac{1}{3}\) rd of \(\mathrm{AB}\) is dissociated then \(\frac{\mathrm{P}}{\mathrm{K}_{\mathrm{p}}}\) will be numerically equal to

5 step solution

Problem 185

For the following reaction in gaseous phase \(\mathrm{CO}+1 / 2 \mathrm{O}_{2} \rightleftharpoons \mathrm{CO}_{2}\) \(\mathrm{K}_{c} / \mathrm{K}_{\mathrm{p}}\) is (a) \((\mathrm{RT})^{1 / 2}\) (b) \((\mathrm{RT})^{-1 / 2}\) (c) (RT) (d) \((\mathrm{RT})^{-1}\)

4 step solution

Problem 186

One of the following equilibria is not affected by change in volume of the flask: (a) \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g}) \mathrm{Cl}_{2}(\mathrm{~g})\) (b) \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\) (c) \(\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\) (d) \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\)

3 step solution

Problem 187

At \(700 \mathrm{~K}\), the equilibrium constant \(\mathrm{K}_{\text {s }}\) for the reaction \(2 \mathrm{SO}_{3}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})\) is \(1.80 \times 10^{-3} .\) What is the numerical value in mole per litre of equilibrium constant \(K_{c}\) for this reaction at the same temperature? \([\mathbf{2 0 0 2}]\) (a) \(8.1 \times 10^{-3}\) (b) \(9.1 \times 10^{-9} \mathrm{~mol} \mathrm{~L}^{-1}\) (c) \(3.1 \times 10^{-7}\) (d) \(6.1 \times 10^{-7} \mathrm{~mol} \mathrm{~L}^{-1}\)

4 step solution

Problem 188

Consider the reaction equilibrium, \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g}) ; \Delta \mathrm{H}^{\circ}=-198 \mathrm{~kJ}\) on the basis of Le Chatelier's principle, the condition favourable for the forward reaction is (a) lowering of temperature as well as pressure (b) increasing temperature as well as pressure (c) lowering the temperature and increasing the pressure (d) any value of temperature and pressure

4 step solution

Problem 189

For the reaction equilibrium, \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\) the concentrations of \(\mathrm{N}_{2} \mathrm{O}_{4}\) and \(\mathrm{NO}_{2}\) at equilibrium are \(4.8 \times 10^{-2}\) and \(1.2 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}\) respectively. The value of \(\mathrm{K}_{\mathrm{c}}\) for the reaction is (a) \(3.3 \times 10^{2} \mathrm{~mol} \mathrm{~L}^{-1}\) (b) \(3 \times 10^{-1} \mathrm{~mol} \mathrm{~L}^{-1}\) (c) \(3 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}\) (d) \(3 \times 10^{3} \mathrm{~mol} \mathrm{~L}^{-1}\)

4 step solution

Problem 190

For the reaction \(\mathrm{CO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightleftharpoons-\mathrm{COCl}_{2}(\mathrm{~g})\), the value of \(\mathrm{Kp} / \mathrm{K}_{c}\) is equal to [2004] (a) RT (b) \(1 / \mathrm{RT}\) (c) I (d) \(\sqrt{\mathrm{RT}}\)

4 step solution

Problem 191

What is the equilibrium expression for the reaction \([\mathbf{2 0 0 4}]\) \(\mathrm{P}_{4}(\mathrm{~s})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons=\mathrm{P}_{4} \mathrm{O}_{10}(\mathrm{~s}) ?\) (a) \(\mathrm{K}_{\mathrm{c}}=\frac{\left[\mathrm{P}_{4} \mathrm{O}_{1}\right]}{5\left[\mathrm{P}_{4}\right]\left[\mathrm{O}_{2}\right]^{5}}\) (b) \(\mathrm{K}_{\mathrm{c}}=\frac{1}{\left[\mathrm{O}_{2}\right]^{5}}\) (c) \(\mathrm{K}_{\mathrm{c}}=\frac{\left[\mathrm{P}_{4} \mathrm{O}_{10}\right]}{\left[\mathrm{P}_{4}\right]\left[\mathrm{O}_{2}\right]^{5}}\) (d) \(\mathrm{K}_{\mathrm{c}}=\left[\mathrm{O}_{2}\right]^{5}\)

4 step solution

Problem 192

The equilibrium constant for the reaction, [2004] \(\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\) At temperature \(\mathrm{T}\) is \(4 \times 10^{-4} .\) The value of \(\mathrm{K}_{\text {cor the }}\) reaction \(\mathrm{NO}(\mathrm{g}) \rightleftharpoons 1 / 2 \mathrm{~N}_{2}(\mathrm{~g})+1 / 2 \mathrm{O}_{2}(\mathrm{~g})\) at the same temperature is (a) \(4 \times 10^{-6}\) (b) \(2.5 \times 10^{2}\) (c) \(0.02\) (d) 50

4 step solution

Problem 193

Consider an endothermic reaction \(\mathrm{X} \longrightarrow \mathrm{Y}\) with the activation energies \(E_{b}\) and \(E_{f}\) for the backward and forward reactions, respectively. In general \(\quad[\mathbf{2 0 0 5}]\) (a) \(\mathrm{E}_{\mathrm{b}}<\mathrm{E}_{\mathrm{f}}\) (b) \(\mathrm{E}_{\mathrm{b}}>\mathrm{E}_{\mathrm{f}}\) (c) \(\mathrm{E}_{\mathrm{b}}=\mathrm{E}_{\mathrm{f}}\) (d) there is no definite relation between \(E_{b}\) and \(E_{f}\)

4 step solution

Problem 194

For the reaction, \(2 \mathrm{NO}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g})\) \(\left(\mathrm{K}_{\mathrm{c}}=1.8 \times 10^{-6}\right.\) at \(\left.184^{\circ} \mathrm{C}\right)\) \((\mathrm{R}=0.0831 \mathrm{~kJ} /(\operatorname{mol} \mathrm{K}))\) when \(K_{\text {and }} K_{c}\) are compared at \(184^{\circ} \mathrm{C}\) it is found that (a) \(\mathrm{K}_{\mathrm{p}}\) is greater than \(\mathrm{K}_{c}\) (b) \(\mathrm{K}_{\mathrm{p}}\) is less than \(\mathrm{K}_{\mathrm{c}}\) (c) \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{c}\) (d) whether \(\mathrm{K}_{p}\) is greater than, less than or equal to \(\mathrm{K}_{\mathrm{c}}\) depends upon the total gas pressure

4 step solution

Problem 195

An amount of solid \(\mathrm{NH}_{4} \mathrm{HS}\) in placed in a flask already containing ammonia gas at a certain temperature and \(0.50\) atm pressure. Ammonium hydrogen sulphide decomposes to yield \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{~S}\) gases in the flask. When the decomposition reaction reaches equilibrium, the total pressure is the flask rises to \(0.84 \mathrm{~atm}\), the equilibrium constant for \(\mathrm{NH}_{4} \mathrm{HS}\) decomposition at this temperature is [2005] (a) \(0.30\) (b) \(0.18\) (c) \(0.17\) (d) \(0.11\)

6 step solution

Problem 196

The exothermic formation of \(\mathrm{ClF}_{3}\) is represented by the equation \([2005]\) \(\mathrm{Cl}_{2}(\mathrm{~g})+3 \mathrm{~F}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{ClF}_{3}(\mathrm{~g}) ; \Delta \mathrm{H}=-329 \mathrm{~kJ}\) Which of the following will increase the quantity of \(\mathrm{CIF}_{3}\) in an equilibrium mixture of \(\mathrm{Cl}_{2}, \mathrm{~F}_{2}\) and \(\mathrm{ClF}_{3} ?\) (a) increasing the temperature (b) removing \(\mathrm{Cl}_{2}\) (c) increasing the volume of the container (d) adding \(\mathrm{F}_{2}\)

7 step solution

Problem 197

Phosphorous pentachloride dissociates as follows, in a closed reaction vessel \(\mathrm{PCI}_{5}(\mathrm{~g}) \longrightarrow \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\) If total pressure at equilibrium of the reaction mixture is \(\mathrm{P}\) and degree of dissociation of \(\mathrm{PCl}_{5}\) is \(\mathrm{x}\), the partial pressure of \(\mathrm{PCl}_{3}\) will be (a) \(\left(\frac{x}{(x+1)}\right) \mathrm{P}\) (b) \(\left(\frac{2 x}{(x-1)}\right) \mathrm{P}\) (c) \(\left(\frac{x}{(x-1)}\right) \mathrm{P}\) (d) \(\left(\frac{x}{(1-x)}\right) \mathrm{P}\)

6 step solution

Problem 198

The equilibrium constant for the reaction \(\mathrm{SO}_{3}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{2}(\mathrm{~g})+1 / 2 \mathrm{O}_{2}(\mathrm{~g})\) is \(\mathrm{K}=4.9 \times 10^{-2}\). the value of \(\mathrm{K}\) for the reaction \(2 \mathrm{SO}_{2}\) \((\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})\) will be \([2006]\) (a) 416 (b) \(2.40 \times 10^{-3}\) (c) \(9.8 \times 10^{-2}\) (d) \(4.9 \times 10^{-2}\)

5 step solution

Problem 199

The equilibrium constants \(K_{p_{1}}\) and \(K_{p_{2}}\) for the reactions \(\mathrm{X} \rightleftharpoons 2 \mathrm{Y}\) and \(\mathrm{Z} \rightleftharpoons \mathrm{P}+\mathrm{Q}\), respectively are in the ratio of \(1: 9\). If the degree of dissociation of \(X\) and \(Z\) be equal then the ratio of total pressure at these equilibria is (a) \(1: 36\) (b) \(1: 1\) (c) \(1: 3\) (d) \(1: 9\)

5 step solution

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Chapter 7 - A Complete Resource Book in Chemistry for JEE Main Solutions — Page 3 | StudyQuestionHub