Problem 80
Question
The ratio of \(\mathrm{Kp} / \mathrm{Kc}\) for the reaction \(\mathrm{SO}_{2}(\mathrm{~g})+1 / 2 \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{3}(\mathrm{~g})\) is (a) \((\mathrm{RT})^{-1 / 2}\) (b) \((\mathrm{RT})^{1 / 2}\) (c) \(\mathrm{RT}\) (d) 1
Step-by-Step Solution
Verified Answer
The ratio \( \frac{K_p}{K_c} \) is \( (RT)^{-1/2} \), corresponding to option (a).
1Step 1: Identify Variables in Ideal Gas Constant Equation
The ratio \( \frac{K_p}{K_c} \) for a gaseous reaction is given by the formula \( K_p = K_c (RT)^{\Delta n} \), where \( \Delta n \) is the change in the number of moles of gas (products - reactants). Here, \( R \) is the ideal gas constant and \( T \) is the temperature in Kelvin.
2Step 2: Compute Delta n (Change in Moles)
For the reaction \( \mathrm{SO_2(g)} + \frac{1}{2}\mathrm{O_2(g)} \rightleftharpoons \mathrm{SO_3(g)} \), calculate \( \Delta n = \text{moles of gaseous products} - \text{moles of gaseous reactants} = 1 - \left( 1 + \frac{1}{2} \right) = 1 - 1.5 = -0.5 \).
3Step 3: Substitute Delta n into the Ratio Formula
Using \( \Delta n = -0.5 \), substitute this into the equation \( \frac{K_p}{K_c} = (RT)^{\Delta n} \) to get \( \frac{K_p}{K_c} = (RT)^{-0.5} \).
4Step 4: Determine Correct Answer Choice
Compare the derived expression \( (RT)^{-0.5} \) with the provided options. The option that matches is (a) \( (RT)^{-1/2} \).
Key Concepts
Equilibrium ConstantsGas LawsReaction QuotientsIdeal Gas Constant
Equilibrium Constants
In chemical reactions, especially those involving gases, understanding equilibrium constants is crucial. These constants help predict how a reaction will proceed under certain conditions. There are two main types:
- \(K_c\): The equilibrium constant for gases expressed in concentrations. It is calculated using molar concentrations of the reactants and products at equilibrium.
- \(K_p\): The equilibrium constant for gases expressed in partial pressures. It is used when dealing with reactions carried out in the gas phase.
Gas Laws
The behavior of gases under different conditions is described by gas laws. These laws are fundamental for understanding reactions involving gases.
- Boyle’s Law: This law states that pressure and volume are inversely proportional, assuming temperature and the number of gas moles remain constant. It can be expressed as \(P_1V_1 = P_2V_2\).
- Charles's Law: Volume and temperature are directly proportional, assuming pressure and the number of moles remain constant. Expressed as \(V_1/T_1 = V_2/T_2\).
- Avogadro’s Law: Volume and the number of moles of the gas are directly proportional at a constant temperature and pressure. It is given by \(V \propto n\).
- Ideal Gas Law: Combines all the simple gas laws into one equation: \(PV = nRT\).
Reaction Quotients
The reaction quotient, \(Q\), is a pivotal concept when analyzing reactions. It's a measure for the relative quantities of products and reactants during a reaction at any given point in time:
- If \(Q = K\), the system is at equilibrium.
- If \(Q < K\), the forward reaction is favored, meaning more products will form.
- If \(Q > K\), the reverse reaction is favored, meaning the reaction will proceed to form more reactants.
Ideal Gas Constant
The ideal gas constant, often symbolized as \(R\), is a crucial parameter in the Ideal Gas Law. It serves to bridge various units used in gas calculations, and its value depends on the units of pressure, volume, and temperature:
- The most common value of \(R\) is 8.314 J/(mol·K), used when pressure is in pascals and volume in cubic meters.
- In contexts where pressure is in atmospheres, its value is taken as 0.0821 L·atm/(mol·K).
Other exercises in this chapter
Problem 75
For the reaction \(\mathrm{C}_{(\mathrm{s})}+\mathrm{CO}_{2}(\mathrm{~g})=2 \mathrm{CO}(\mathrm{g})\), the par- tial pressures of \(\mathrm{CO}_{2}\) and \(\mat
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In which of the following gaseous reaction, the value of \(K_{p}\) is less than \(K_{c}\) ? (a) \(\mathrm{PCl}_{5} \rightleftharpoons \mathrm{PCI}_{3}+\mathrm{C
View solution Problem 81
If \(2 \mathrm{NO} \rightleftharpoons \mathrm{N}_{2}+\mathrm{O}_{2}\) \(\mathrm{K}_{\mathrm{c}}=2.5 \times 10^{30}\) \(\mathrm{NO}+\frac{1}{2} \mathrm{Br}_{2} \
View solution Problem 84
\(9.2\) grams of \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})\) is taken in a closed one litre vessel and heated till the following equilibrium is reached \(\ma
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