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Byju's Answer

Standard XII

Chemistry

First Order Reaction

For a first o...

Question

For a first order reaction involving decomposition of $N_{2} O_{5}$ the following information is available
$2 N_{2} O_{5} (g) \to 4 N O_{2} (g) + O_{2} (g)$ rate $= k [N_{2} O_{5}]$
$N_{2} O_{5} (g) \to 2 N O_{2} (g) + 1 / 2 O_{2} (g)$ rate $= k^{'} [N_{2} O_{5}]$

k = k^{'}

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k^{'} = 2 k

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k^{'} = 1 / 2 k

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k > k^{'}

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Solution

The correct option is D $k > k^{'}$
${2 N_{2} O_{5}}_{(g)} ⟶ {4 N O_{2}}_{(g)} + {O_{2}}_{(g)}$
$\Rightarrow k = \frac{[N O_{2}]^{4} [O_{2}]}{[N_{2} O_{5}]^{2}} ⟶ (1)$
${N_{2} O_{5}}_{(g)} ⟶ {2 N O_{2}}_{(g)} + {1 / 2 O_{2}}_{(g)}$
$\Rightarrow k^{'} = \frac{[N O_{2}]^{2} [O_{2}]^{1 / 2}}{[N_{2} O_{5}]}$
$\Rightarrow [N_{2} O_{5}] = \frac{[N O_{2}]^{2} [O_{2}]^{1 / 2}}{k^{'}} ⟶ (2)$
$(2)$ in $(1) \Rightarrow k = \frac{[N O_{2}]^{4} [O_{2}]}{\frac{[N O_{2}]^{4} [O_{2}]}{k^{' 2}}} = k^{' 2}$
$\Rightarrow k > k^{'}$

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Similar questions

Q. For the first order decomposition reaction of

N_{2} O_{5}

, it is observed that

(i) N_{2} O_{5} (g) \to 2 N O_{2} (g) + \frac{1}{2} O_{2} (g); \frac{- d [N_{2} O_{5}]}{d t} = k [N_{2} O_{5}] (i i) 2 N_{2} O_{5} (g) \to 4 N O_{2} (g) + O_{2} (g); \frac{- d [N_{2} O_{5}]}{d t} = k^{'} [N_{2} O_{5}]

Which of the following is true?

Q. The rate of the reaction

2 N_{2} O_{5} \to 4 N O_{2} + O_{2}

can be written in three ways:

\frac{- d [N_{2} O_{5}]}{d t} = k [N_{2} O_{5}]

\frac{d [N O_{2}]}{d t} = k^{^{'}} [N_{2} O_{5}]

\frac{d [O_{2}]}{d t} = k^{''} [N_{2} O_{5}]

The relationship between

k

k^{^{'}}

and between

k

k^{''}

are:

Q. The decomposition of

N_{2} O_{5}

at 318 K according to the following equation follows first order reaction:

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

The initial concentration of

N_{2} O_{5}

was

1.24 \times 10^{- 2}

mol

L^{- 1}

and after 60 minutes was

0.20 \times 10^{- 2}

L^{- 1}

, Calculate the rate constant of the reaction at 318 K.

Q. The rate constant,

k

for the reaction

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

2.3 \times 10^{- 2} s^{- 1}

. Which equation given below describes the change of

[N_{2} O_{5}]

with time?

[N_{2} O_{5}]_{0}

and

[N_{2} O_{5}]_{t}

correspond to concentration of

N_{2} O_{5}

initially and at time

t

Q. For a first order reaction involving decomposition of

N_{2} O_{5}

, the following information is available :

2 N_{2} O_{5} (g)

\to

{4 N O}_{2} (g)

O_{2} (g)

Rate = k

[N_{2} O_{5}]

N_{2} O_{5}

(g)

\to

{2 N O}_{2}

(g) +

\frac{1}{2} O_{2}

(g) Rate =

k^{'}

[N_{2} O_{5}]

,
Which of the following expressions is true ?

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For a first order reaction involving decomposition of N2O5 the following information is available2N2O5(g)→4NO2(g)+O2(g) rate =k[N2O5]N2O5(g)→2NO2(g)+1/2O2(g) rate =k′[N2O5]

The correct option is D k>k′2N2O5(g)⟶4NO2(g)+O2(g)⇒k=[NO2]4[O2][N2O5]2⟶(1)N2O5(g)⟶2NO2(g)+1/2O2(g)⇒k′=[NO2]2[O2]1/2[N2O5]⇒[N2O5]=[NO2]2[O2]1/2k′⟶(2)(2) in (1)⇒k=[NO2]4[O2][NO2]4[O2]k′2=k′2⇒k>k′

For a first order reaction involving decomposition of $N_{2} O_{5}$ the following information is available
$2 N_{2} O_{5} (g) \to 4 N O_{2} (g) + O_{2} (g)$ rate $= k [N_{2} O_{5}]$
$N_{2} O_{5} (g) \to 2 N O_{2} (g) + 1 / 2 O_{2} (g)$ rate $= k^{'} [N_{2} O_{5}]$