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Factors Affecting the Rate of a Chemical Reaction

The reaction ...

Question

The reaction $2 H_{2 (g)} + 2 N O_{(} g) \to 2 H_{2} O_{(} g) + N_{2 (g)}$ is first order in NO. The rate constant of the reaction at a certain temperature is $4.42 M^{- 2} s^{- 1}$ Calculate the rate when $[H_{2}] = 0.015 M$ and [NO]= 0.025 M ?

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Solution

The given reaction is:

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

According to questions, the reaction is first order with respect to $N O$ . So,
$R a t e = K [N O]$
where, $K$ is a rate constant.

Now, $K = 4.42 M^{- 2} s^{- 1}$ , $[H_{2}] = 0.015 M$ , $[N O] = 0.025 M$
Putting these values in $e q^{n} (i)$ :-
Rate= $4.42 \times 0.025 M s^{- 1}$
= $0.1105 M s^{- 1}$

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Similar questions

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

H_{2}

0.42 M^{- 2} s^{- 1}

[H_{2}] = 0.015 M

0.025

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

H_{2}

N O

M^{- 2} S^{- 1}

[H_{2}]

N O

=

2 H_{2}

\to

_{2}

_{2}

H_{2}

M^{- 2}

s^{- 1}

[H_{2}]

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

\frac{d [N_{2}]}{d t} = k [H_{2}] [N O]^{2}

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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