Question

Hydrogen reacts with nitrogen monoxide to form dinitrogen monoxide (laughing gas) according to the equation: H2O(g) + 2NO(g) rightarrow O + H2O(g)NO (M) 0.30 0.60 0.60H2 (M) 0.35 0.35 0.70Rate (M/L/s) 3.822 x 103 1.529 x 10-2 3.058 x 10-2 A) Determine the rate law according to the following data. B) Determine the rate constant (in mol-2. L2. s-1).C) Determine the orders with respect to each reactant.

Answer 1

Answer:

A. $r=k[NO]^2[H_2]$

B. $k=0.121\frac{L^2}{mol^2*s}$

C. Second-order with respect to NO and first-order with respect to H₂

Explanation:

Hello,

In this case, for the reaction:

$H_2O(g) + 2NO(g) \rightarrow N_2O + H_2O(g)$

The rate law is determined by writing the following hypothetical rate laws:

$3.822x10^{-3}=k[0.3]^m[0.35]^n\\\\1.529x10^{-2}=k[0.6]^m[0.35]^n\\\\3.058x10^{-2}=k[0.6]^m[0.7]^n$

Whereas we can compute m as follows:

$\frac{3.822x10^{-3}}{1.529x10^{-2}} =\frac{[0.3]^m[0.35]^n}{[0.6]^m[0.35]^n} \\\\0.25=(0.5)^m\\\\m=\frac{log(0.25)}{log(0.5)} \\\\m=2$

Therefore, the reaction is second-order with respect to NO. Thus, for hydrogen, we find n:

$\frac{1.529x10^{-2}}{3.058x10^{-2}} =\frac{[0.6]^2[0.35]^n}{[0.6]^2[0.7]^n} \\\\0.5=(0.5)^n\\\\n=\frac{log(0.5)}{log(0.5)}\\ \\n=1$

A) Therefore, the reaction is first-order with respect to H₂. In such a way, we conclude that that the rate law is:

$r=k[NO]^2[H_2]$

B) Rate constant is computed from one kinetic data:

$k=\frac{1.529x10^{-2}\frac{mol}{L*s} }{(0.6\frac{mol}{L} )^2(0.35\frac{mol}{L})}\\\\k=0.121\frac{L^2}{mol^2*s}$

C. As mentioned before, reaction is second-order with respect to NO and first-order with respect to H₂.

Best regards.