The reaction $A + 2 B \to C + D$ , obey the rate of reaction, Rate $= K A^{x} B^{y}$ . What should be the order of reaction. $2 N_{2} O_{5} (g) \to 4 N O_{2} (g) + O_{2} (g)$ .

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Solution

Given reaction is
$2 N_{2} O_{5} ⟶ 4 N O_{2} + O_{2}$
Let rate of reaction be
$R a t e_{1} = k {[N_{2} O_{5}]}^{2}$
When concentration of $N_{2} O_{5}$ is doubled
$R a t e_{2} = k {[{2 N}_{2} O_{5}]}^{2} = 4 k {[N_{2} O_{5}]}^{2} = 4 R a t e_{1}$
On tripling the concentration
$R a t e_{3} = k {[3 N_{2} O_{5}]}^{2} = 9 k {[N_{2} O_{5}]}^{2} = 9 R a t e_{1}$
Since reaction rate is Proportional to ${[N_{2} O_{5}]}^{2},$ The reaction is dearly a second order reaction.

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Similar questions

2 N_{2} O_{5 (g)} \to 4 N O_{2 (g)} + O_{2 (g)}

2 N_{2} O_{5} (g) \to 4 N O_{2} (g) + O_{2} (g)

N_{2} O_{5}

2 N_{2} O_{5} (g) \to 4 N O_{2} (g) + O_{2} (g)

= k [N_{2} O_{5}]

N_{2} O_{5} (g) \to 2 N O_{2} (g) + 1 / 2 O_{2} (g)

= k^{'} [N_{2} O_{5}]

2 N_{2} O_{5} (g) \to 4 N O_{2} (g) + O_{2} (g)

N O_{2}

2.8 \times 10^{- 3} M s^{- 1}

N_{2} O_{5} (g)

For the following reaction,

$2 N_{2} O_{5} (g)$ $\to$ $4 N O_{2} (g) + O_{2} (g)$

State the rate expression.

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