Q17.82CP

Question

Isolation of Group $8 B (10)$ elements, used as industrial catalysts, involves a series of steps. For nickel, the sulfide ore is roasted in air: ${Ni}_{3} S_{2} (s) + O_{2} (g) \to NiO (s) + {SO}_{2} (g)$ . The metal oxide is reduced by the $H_{2}$ in water gas $(CO + H_{2})$ to impure $Ni : NiO (s) + H_{2} (g) \to Ni (s) + H_{2} O (g)$ . The CO in water gas then reacts with the metal in the Mond process to form gaseous nickel carbonyl, $Ni (s) + CO (g) \to Ni {(CO)}_{4} (g)$ , which is subsequently decomposed to the metal.

(a) Balance each of the three steps, and obtain an overall balanced equation for the conversion of ${Ni}_{3} S_{2}$ to $Ni {(CO)}_{4}$ .

(b) Show that the overall $Q_{c}$ is the product of the $Q_{c}$ 's for the individual reactions.

Step-by-Step Solution

Verified

Answer

(a) The equation is:

$2 {Ni}_{3} S_{2} (s) + 7 O_{2} (g) + 6 H_{2} (g) + 24 CO (g) \to 6 Ni {(CO)}_{4} (g) + 4 {SO}_{2} (g) + 6 H_{2} O (g)$

(b) $Q_{0 n} = Q_{c}$

The reaction quotient for the overall reaction is equal to the product of the reaction quotients for the each step

1Step 1: Definition of Concept

Reactions: A balanced chemical reaction equation shows the mole relationships of reactants and products, while a chemical reaction equation gives the reactants and products. The amount of energy involved in the reaction is frequently stated. Reaction stoichiometry is the study of the quantitative aspects of chemical reactions.

2Step 2: Find the overall balanced equation

(a)

Considering the given information:

${Ni}_{3} S_{2} (s) + O_{2} (g) \to NiO (s) + {SO}_{2} (g) NiO (s) + H_{2} (g) \to Ni (s) + H_{2} O (g) Ni (s) + CO (g) \to Ni {(CO)}_{4} (g)$

We can start by balancing the nickel and sulphur in reaction 1.

${Ni}_{3} S_{2} (s) + O_{2} (g) \to 3 NiO (s) + 2 {SO}_{2} (g)$

After that, you must restore the oxygen balance.

${Ni}_{3} S_{2} (s) + \frac{7}{2} O_{2} (g) \to 3 NiO (s) + 2 {SO}_{2} (g)$

To make the coefficients integer, multiply everything by two.

$2 {Ni}_{3} S_{2} (s) + 7 O_{2} (g) \to 6 NiO (s) + 4 {SO}_{2} (g)$

Because $6 NiO (s)$ is used in reaction (2), we must also react $6 NiO (s)$ . Because everything appears to be in balance, we must make each coefficient 6 as well.

$6 NiO (s) + 6 H_{2} (g) \to 6 Ni (s) + 6 H_{2} O (g)$

Because there are $6 Ni (s)$ produced in reaction (3), we must react 6 moles of nickel in the following reaction.

$Ni (s) + CO (g) \to Ni {(CO)}_{4} (g)$

Make sure the rest of the elements are in order.

$6 Ni (s) + 24 CO (g) \to 6 Ni {(CO)}_{4} (g)$

3Step 3: Show that the overall Q c is the product of the Q c 's for the individual reactions.

(b)

Write the overall reaction first.

$2 {Ni}_{3} S_{2} (s) + 7 O_{2} (g) \to 6 NiO (s) + 4 {SO}_{2} (g) 6 NiO (s) + 6 H_{2} (g) \to 6 Ni (s) + 6 H_{2} O (g) 6 Ni (s) + 24 CO (g) \to 6 Ni {(CO)}_{4} (g) 2 {Ni}_{3} S_{2} (s) + 7 O_{2} (g) + 6 H_{2} (g) + 24 CO (g) \to 4 {SO}_{2} (g) + 6 Ni {(CO)}_{4} (g) + 6 H_{2} O (g)$

In terms of concentration, write the expression for the overall reaction's equilibrium constant.

$Q_{c} = \frac{[product]}{[reactant]} Q_{c} = \frac{{[{SO}_{2}]}^{4} {[Ni {(CO)}_{4}]}^{6} {[H_{2} O]}^{6}}{{[O_{2}]}^{7} {[H_{2}]}^{6} {[CO]}^{24}}$

In terms of concentration, write the expression for the equilibrium constant of reaction (1).

$2 {Ni}_{3} S_{2} (s) + 7 O_{2} (g) \to 6 Ni (s) + 4 {SO}_{2} (g) Q_{c 1} = \frac{[product]}{[reactant]} Q_{c 1} = \frac{{[{SO}_{2}]}^{4}}{{[O_{2}]}^{7}}$

In terms of concentration, write the expression for the equilibrium constant of reaction (2).

$6 NiO (s) + 6 H_{2} (g) \to 6 Ni (s) + 6 H_{2} O (g) Q_{c 2} = \frac{[product]}{[reactant]} Q_{c 2} = \frac{{[H_{2} O]}^{6}}{{[H_{2}]}^{6}}$

In terms of concentration, write the expression for the equilibrium constant of reaction (3).

$6 Ni (s) + 24 CO (g) \to 6 Ni {(CO)}_{4} (g) Q_{c 3} = \frac{[product]}{[reactant]} Q_{c 3} = \frac{{[Ni {(CO)}_{4}]}^{6}}{{[CO]}^{24}}$

Show that $Q_{c} = Q_{c 1} \times Q_{c 2} \times Q_{c 3}$

$Q_{c} = Q_{c 1} \times Q_{c 2} \times Q_{c 3} \frac{{[{SO}_{2}]}^{4} {[H_{2} O]}^{6} {[Ni {(CO)}_{4}]}^{6}}{{[O_{2}]}^{7} {[H_{2}]}^{6} {[CO]}^{24}} = \frac{{[{SO}_{2}]}^{4}}{{[O_{2}]}^{7}} \times \frac{{[H_{2} O]}^{6}}{{[H_{2}]}^{6}} \times \frac{{[Ni {(CO)}_{4}]}^{6}}{{[CO]}^{24}} {[O_{2}]}^{7} {[H_{2}]}^{6} {[CO]}^{24} = {[O_{2}]}^{7} {[H_{2}]}^{6} {[CO]}^{24}$

As observed, $Q_{c} = Q_{c 1} \times Q_{c 2} \times Q_{c 3}$ .

Q17.79CP

Q17.83CP

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