The reaction: 2NO+2H2⟶N2+2H2O has been assigned to follow the following mechanism1. NO+NO⇌N2O2 (fast)2. N2O2+H2⟶N2O+H2O (slow)3. N2O+H2⟶N2+H2O (fast)The rate constant of step 2 is 1.2×10−4mole−1L min-1while equilibrium constant of step 1 is 1.4×10−2 What is the rate of reaction when concentration of NO and H2 each is 0.5 mole L −1 ?

3.5 × 10 − 4 mol L − 1 min − 1

3.5 × 10 − 4 mol L − 1 min − 1

2.1 × 10 − 7 mol L − 1 min − 1

3.2 × 10 − 6 mol L − 1 min − 1

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The reaction: $2 NO + 2 H_{2} ⟶ N_{2} + 2 H_{2} O$ has been assigned to follow the following mechanism
1. $NO + NO ⇌ N_{2} O_{2}$ (fast)
2. $N_{2} O_{2} + H_{2} ⟶ N_{2} O + H_{2} O$ (slow)
3. $N_{2} O + H_{2} ⟶ N_{2} + H_{2} O$ (fast)
The rate constant of step 2 is $1.2 \times 10^{- 4} {mole}^{- 1} L \min^{- 1}$ while equilibrium constant of step 1 is $1.4 \times 10^{- 2}$ What is the rate of reaction when concentration of NO and $H_{2}$ each is $0.5 {mole L}^{- 1} ?$

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$3.5 \times 10^{- 4} mol L - 1 \min^{- 1}$

$2.1 \times 10^{- 7} mol L^{- 1} \min^{- 1}$

None of the above

$3.2 \times 10^{- 6} mol L^{- 1} \min^{- 1}$

answer is A.

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$r = k \cdot [H_{2}] [N_{2} O_{2}]$

Also, $[N_{2} O_{2}] = k_{e q} \cdot [NO]^{2}$

$\Rightarrow r = k \cdot k_{e q} \cdot [H_{2}] [NO]^{2}$

$\begin{array}{l} r = 1.2 \times 10^{- 4} \times 1.4 \times 10^{- 2} \times (0.5)^{3} \\ = 2.1 \times 10^{- 7} mol L^{- 1} \min^{- 1} \end{array}$

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