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Methods for Ester Synthesis

For the react...

Question

For the reaction

$2 H_{2} (g) + 2 N O (g) \to N_{2} (g) + 2 H_{2} O (g)$

the observed rate expression is,

rate = $k_{f} [N O]^{2} [H_{2}]$ . The rate expression for the reverse reaction is :

k_{b} \frac{[N_{2}] [H_{2} O]^{2}}{[H_{2}]}

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k_{b} \frac{[N_{2}] [H_{2} O]^{2}}{[N O]^{2}}

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k_{b} [N_{2}] [H_{2} O]^{2}

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k_{b} [N_{2}] [H_{2} O]

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Solution

The correct option is A $k_{b} \frac{[N_{2}] [H_{2} O]^{2}}{[H_{2}]}$
$k_{e q} = \frac{k_{f}}{k_{b}} = \frac{[N_{2}] [H_{2} O]^{2}}{[H_{2}]^{2} [N O]^{2}}$

Rearranging

$k_{f} [N O]^{2} [H_{2}] = \frac{k_{b} [N_{2}] [H_{2} O]^{2}}{[H_{2}]}$

on comparing $R_{f}$ and $R_{b}$ at equilibrium,

$R_{b} = k_{b} \frac{[N_{2}] [H_{2} O]^{2}}{[H_{2}]}$

Suggest Corrections

Similar questions

2 H_{2} (g) + 2 N O (g) \to N_{2} (g) + 2 H_{2} O (g)

= k_{f} [N O]^{2} [H_{2}]

2 N O (g) + 2 H_{2} (g) \to N_{2} (g) + 2 H_{2} O (g)

\frac{d [N_{2}]}{d t} = k_{1} [N O] [H_{2}]; \frac{d [H_{2} O]}{d t} = k_{2} [N O] [H_{2}]; \frac{- d [N O]}{d t} = k_{3} [N O] [H_{2}]; \frac{d [H_{2}]}{d t} = k_{4} [N O] [N_{2}]

k_{1}, k_{2}, k_{3}, k_{4}

2 N O (g) + 2 H_{2} (g) \to N_{2} (g) + 2 H_{2} O (g)

\frac{d [N_{2}]}{d t} = k_{1} [N O] [H_{2}] \frac{d [H_{2} O]}{d t} = k [N O] [H_{2}] - \frac{d [N O]}{d t} = k_{1}^{'} [N O] [H_{2}] - \frac{d [H_{2}]}{d t} = k_{1}^{''} [N O] [H_{2}]

k, k_{1}, k_{1}^{'}

k_{1}^{''}

2 N O (g) + 2 H_{2} (g) \to N_{2} (g) + 2 H_{2} O (g)

\frac{d [N_{2}]}{d t} = k_{1} [N O] [H_{2}] \frac{d [H_{2} O]}{d t} = k [N O] [H_{2}] - \frac{d [N O]}{d t} = k_{1}^{'} [N O] [H_{2}] - \frac{d [H_{2}]}{d t} = k_{1}^{''} [N O] [H_{2}]

k, k_{1}, k_{1}^{'}

k_{1}^{''}

2 H_{2} (g) + 2 N O (g) \to N_{2} (g) + 2 H_{2} O (g)

R a t e = k [H_{2}] [N O]^{2}

2 N O (g) ⇌ N_{2} O_{2} (g)

N_{2} O_{2} + H_{2} \to N_{2} O + H_{2} O

N_{2} O + H_{2} \to H_{2} O + N_{2}

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