Question

14. The following reaction mechanisms are used for the final reaction shown: 203 → 302 Step 1: CI+O3 → 02 + Clo Step 2: 03 → 02 +0 Step 3: ClO +002+CI a. What is the catalyst in the reaction? b. What is/are the intermediate product(s) for the reaction? 15. At a particular moment during the reaction, Hl disappears at a rate of 6.50 X 102 M/s. What is the rate of appearance of H2 and the rate of appearance of 12 2 HI() →H2(g)+12(e) 16. Write the equilibrium expressions for the following reactions: a. 2 NO(g) → N2(g) + O2(B) b. 2 Fe (s) + 3 H2O(6) ► Fe2O3 (s) + 3H2 (8) 17. The Kp value for the following reaction is 0.56 at 1024°C. What is the Kc value at this temperature? H2 (6) + 12 (g) → 2 HI (B)

Answer 1

14) a) Cl (free radical) is catayst in the reaction as it helps in the initiation of decomposition of ozone .

(03

b) Species which are not present in the in the main reaction termed as intermediate product therfore, ClO and O are intermediate products.

15) Given reaction :

2HIg) + H2(g) + L2(g)

Rate of disappearance of HI $\mathrm{[-\frac{\mathrm{d} [HI]}{\mathrm{d} t}]}$ is  $\mathrm{6.50\times 10^{-1}\ M/s}$.

Relation between rate of disappearance of HI and rate of appearance of is shown below.

H

$\mathrm{-\frac{1}{2}\frac{\mathrm{d} [HI]}{\mathrm{d} t}=\frac{\mathrm{d} [H_{2}]}{\mathrm{d} t}}$

Substituting the value of $\mathrm{-\frac{\mathrm{d} [HI]}{\mathrm{d} t}}$ ,

$\mathrm{\frac{1}{2}\times 6.50\times 10^{-1}\ M/s=\frac{\mathrm{d} [H_{2}]}{\mathrm{d} t}}$

or,

d[12] dt 3.25 x 10-1 M/s

Relation between rate of disappearance of HI and rate of appearance of $\mathrm{I_{2}}$ is shown below.

$\mathrm{-\frac{1}{2}\frac{\mathrm{d} [HI]}{\mathrm{d} t}=\frac{\mathrm{d} [I_{2}]}{\mathrm{d} t}}$

Substituting the value of $\mathrm{-\frac{\mathrm{d} [HI]}{\mathrm{d} t}}$ ,

$\mathrm{\frac{1}{2}\times 6.50\times 10^{-1}\ M/s=\frac{\mathrm{d} [I_{2}]}{\mathrm{d} t}}$

or, $\mathrm{\frac{\mathrm{d} [I_{2}]}{\mathrm{d} t}=\mathbf{3.25\times 10^{-1}\ M/s}}$

Therefore, rate of appearance of
H
;
d[12] dt 3.25 x 10-1 M/s

Therefore, rate of appearance of $\mathrm{I_{2}}$ ; $\mathrm{\frac{\mathrm{d} [I_{2}]}{\mathrm{d} t}=\mathbf{3.25\times 10^{-1}\ M/s}}$

16) Equlibrium expression can be written as shown below.

$\mathrm{K=\frac{[product_{1}]^{stoichiometry\ coefficient}[product_{2}]^{stoichiometry\ coefficient}}{[reactant_{1}]^{stoichiometry\ coefficient}[reactant_{2}]^{stoichiometry\ coefficient}}}$

a) Given reaction,

$\mathrm{2NO(g)\rightarrow N_{2}(g)+O_{2}(g)}$

Equilibrium expression for the above reaction is shown below.

$\mathrm{\mathbf{K=\frac{[N_{2}][O_{2}]}{[NO]^{2}}}}$

b) Given reaction,

$\mathrm{2Fe(s)+3H_{2}O(g)\rightarrow Fe_{2}O_{3}(s)+3H_{2}(g)}$

Equilibrium expression for the above reaction is shown below.

$\mathrm{K=\frac{[Fe]^{2}[H_{2}O]^{3}}{[Fe_{2}O_{3}][H_{2}]^{3}}}$

$\mathrm{[Fe]=[Fe_{2}O_{3}]=1}$ (since present in solid sate)

$\mathrm{\therefore\mathbf{ K=\frac{[H_{2}O]^{3}}{[H_{2}]^{3}}}}$

17) Given reaction :

$\mathrm{H_{2}(g)+I_{2}(g)\rightarrow 2HI(g)}$

Relation between $\mathrm{K_{p}\ and\ K_{c}}$ is shown below.

$\mathrm{K_{p}=K_{c}(RT)^{\Delta n}}$ ----------------------(1)

Given $\mathrm{K_{p}=0.56}$ ; R= gas constant

$\mathrm{\Delta n}$ = (total moles of products)-(total moles of reactants)

= 2 - (1+1)

= 0

Substituting in eq (1),

$\mathrm{0.56=K_{c}(RT)^{0}}$

$\mathrm{\mathbf{K_{c}=0.56}}$

Therefore , for given reaction $\mathrm{\mathbf{K_{c}=0.56}}$ .