Solveeit Logo
Login

Question

Question: For the reaction, 2NO(g) + 2H2(g) \(\longrightarrow\) N2(g) + 2H2O(g) the rate expression can be wri...

For the reaction, 2NO(g) + 2H2(g) ⟶\longrightarrow N2(g) + 2H2O(g) the rate expression can be written in the following ways:

$$$\{\text{- d}\lbrack\text{NO}\rbrack\text{ / dt}\}\text{ = k 1 }\lbrack\text{NO}\rbrack\ \lbrack\text{H2}\rbrack\text{ ; }\{\text{-d}\lbrack\text{H2}\rbrack\text{ / dt}\}\text{ = k 1 }\lbrack\text{NO}\rbrack\lbrack\text{H2}\rbrack$ The relationship between $\text{k, }\text{k}_{1}\text{ , k}\text{'}_{1}\text{ and k'}\text{'}_{1}.$ is :
A

k = k1 = k’ 1 = k’’1\text{k = }\text{k}_{1}\text{ = k'}\text{ }_{1}\text{ = k'}\text{'}_{1}

B

k = 2k1 = k’ 1 = k’’1 \text{k = 2}\text{k}_{1}\text{ = k'}\text{ }_{1}\text{ = k'}\text{'}_{1}\

C

k = 2k’1 = k1 = k’’1\text{k = 2k}\text{'}_{1}\text{ = }\text{k}_{1}\text{ = k'}\text{'}_{1}

D

k = k1 = k’1 = 2 k’’1\text{k = }\text{k}_{1}\text{ = k}\text{'}_{1}\text{ = 2 k'}\text{'}_{1}

Answer

k = 2k1 = k’ 1 = k’’1 \text{k = 2}\text{k}_{1}\text{ = k'}\text{ }_{1}\text{ = k'}\text{'}_{1}\

Explanation

Solution

2NO(g)+2H2(g)→N2(g)+2H2O(g)2NO(g) + 2H_{2}(g)\overset{\quad\quad}{\rightarrow}N_{2}(g) + 2H_{2}O(g)

Rate=−12d[NO]dt=−12d[H2]dt=d[N2]dt=12d[H2O]dt= K1[NO] [H2]Rate = - \frac{1}{2}\frac{d\lbrack NO\rbrack}{dt} = - \frac{1}{2}\frac{d\left\lbrack H_{2} \right\rbrack}{dt} = \frac{d\left\lbrack N_{2} \right\rbrack}{dt} = \frac{1}{2}\frac{d\left\lbrack H_{2}O \right\rbrack}{dt}\text{= }\text{K}_{1}\lbrack\text{NO}\rbrack\ \lbrack H_{2}\rbrack (1)d[H2O]dt=2K1[NO][H2]=K[NO][H2]Sok=2k1\frac{d\left\lbrack H_{2}O \right\rbrack}{dt} = 2K_{1}\lbrack NO\rbrack\left\lbrack H_{2} \right\rbrack = K\lbrack NO\rbrack\left\lbrack H_{2} \right\rbrack So k = 2k_{1}

(2)−d[NO]dt= 2k1[NO][H2] = K1’ [NO] [H2]\mathbf{-}\frac{\mathbf{d}\left\lbrack \mathbf{NO} \right\rbrack}{\mathbf{dt}}\text{= 2}\text{k}_{\mathbf{1}}\mathbf{\lbrack}\text{NO}\mathbf{\rbrack\lbrack}\mathbf{H}_{\mathbf{2}}\mathbf{\rbrack}\text{ = }\text{K}_{\mathbf{1}}\text{' }\mathbf{\lbrack}\text{NO}\mathbf{\rbrack}\mathbf{\ }\mathbf{\lbrack}\mathbf{H}_{\mathbf{2}}\mathbf{\rbrack}

(3) −d[H2]dt= 2k1[NO][H2] = K1" [NO] [H2] \mathbf{-}\frac{\mathbf{d}\left\lbrack \mathbf{H}_{\mathbf{2}} \right\rbrack}{\mathbf{dt}}\text{= 2}\text{k}_{\mathbf{1}}\mathbf{\lbrack}\text{NO}\mathbf{\rbrack\lbrack}\mathbf{H}_{\mathbf{2}}\mathbf{\rbrack}\text{ = }\text{K}_{\mathbf{1}}\text{" }\mathbf{\lbrack}\text{NO}\mathbf{\rbrack}\mathbf{\ }\mathbf{\lbrack}\mathbf{H}_{\mathbf{2}}\mathbf{\rbrack}\mathbf{\ }

k1" = 2K1k_{1}\text{" = 2}\text{K}_{1}