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Q.

Consider the reaction,
$2 H_{2} (g) + 2 NO (g) ⟶ N_{2} (g) + 2 H_{2} O (g)$
The rate law for this reaction will be, if mechanism of reaction is :
$Step 1: NO ⇌ N_{2} O_{2} [fast equilibrium]$
$Step 2: N_{2} O_{2} + H_{2} ⟶ N_{2} O + H_{2} O [slow]$
$Step 3 : N_{2} O + H_{2} ⟶ N_{2} + H_{2} O [fast]$

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a

$r = k [H_{2}] [NO]^{2}$

b

$r = k {[H_{2}]}^{2} [NO]$

c

$r = k {[H_{2}]}^{2} [NO]^{2}$

d

$r = k [H_{2}] [NO]$

answer is C.

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Detailed Solution

$If step (b) is slowest step$

$r = k_{2} [N_{2} O_{2}] [H_{2}]$

$\frac{k_{f}}{k_{b}} = \frac{[N_{2} O_{2}]}{[NO]^{2}}$

$[N_{2} O_{2}] = \frac{k_{f}}{k_{b}} = [{NO}^{2}$

$r = k_{2} = (\frac{k_{f}}{k_{b}}) [H_{2}] [NO]^{2}$

$r = k [H_{2}] [{NO}^{2}$

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Consider the reaction,2H2(g)+2NO(g)⟶N2(g)+2H2O(g)The rate law for this reaction will be, if mechanism of reaction is : Step 1: NO⇌N2O2 [fast equilibrium] Step 2: N2O2+H2⟶N2O+H2O [slow] Step 3:N2O+H2⟶N2+H2O [fast]