For the reaction 2N2O5g - 4NO2g - O2g- - - N2O5 - t- k1 - N2O5 - - - NO2 - t - k2 - N2 O5 - - - - O2 - t - k3 - N2 O5 - Thus-

nbsp;For the reaction nbsp; 2N_2O_5(g) → 4NO_2(g) + O_2(g) Rate of the reaction: 12d[N_2O_5]dt=k_12N_2O_5] (i) 14d[NO_2]dt=k_24[N_2O_5] (ii) d[O_2]dt=k_ ...

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Byju's Answer

Standard XII

Chemistry

Rate of Reaction

For the react...

Question

For the reaction
$2 N_{2} O_{5} (g) \to 4 N O_{2} (g) + O_{2} (g)$
$- \frac{Δ [N_{2} O_{5}]}{Δ t} = k_{1} [N_{2} O_{5}], + \frac{[N O_{2}]}{Δ t} = k_{2} [N_{2} O_{5}], + \frac{Δ [O_{2}]}{Δ t} = k_{3} [N_{2} O_{5}]$ Thus,

2 k_{1} = 4 k_{2} = k_{3}

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2 k_{1} = k_{2} = 4 k_{3}

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k_{1} = k_{2} = k_{3}

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None of these

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Solution

The correct option is B $2 k_{1} = k_{2} = 4 k_{3}$
For the reaction
$2 N_{2} O_{5} (g) \to 4 N O_{2} (g) + O_{2} (g)$
Rate of the reaction:
$- \frac{1}{2} \frac{d [N_{2} O_{5}]}{d t} = \frac{k_{1}}{2} N_{2} O_{5}] (i)$
$\frac{1}{4} \frac{d [N O_{2}]}{d t} = \frac{k_{2}}{4} [N_{2} O_{5}] (i i)$
$\frac{d [O_{2}]}{d t} = k_{3} [N_{2} O_{5}] (i i i)$
all are rate of reaction thus
$i = i i = i i i$
$2 k_{1} = k_{2} = 4 k_{3}$

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Similar questions

Q. For the reaction

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

- \frac{Δ [N_{2} O_{5}]}{Δ t} = k_{1} [N_{2} O_{5}], + \frac{[N O_{2}]}{Δ t} = k_{2} [N_{2} O_{5}], + \frac{Δ [O_{2}]}{Δ t} = k_{3} [N_{2} O_{5}]

Thus,

Q. For a reaction,

N_{2} O_{5} ⟶ 2 N O_{2} + \frac{1}{2} O_{2}

Given:

\frac{- d [N_{2} O_{5}]}{d t} = k_{1} [N_{2} O_{5}]

\frac{d [N O_{2}]}{d t} = k_{2} [N_{2} O_{5}]

\frac{d [O_{2}]}{d t} = k_{3} [N_{2} O_{5}]

The relation between

k_{1}, k_{2}

and

k_{3}

are:

For the reaction;

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

Given:

- \frac{d [N_{2} O_{5}]}{d t} = K_{1} [N_{2} O_{5}], \frac{d [N O_{2}]}{d t} = K_{2} [N_{2} O_{5}], \frac{d [O_{2}]}{d t} = K_{3} [N_{2} O_{5}]

.
The relation between

K_{1}, K_{2}

and

K_{3}

is:

Q. For the reaction

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

, if

- \frac{d [N_{2} O_{5}]}{d t} = k_{1} [N_{2} O_{5}]

\frac{d [N O_{2}]}{d t} = k_{2} [N_{2} O_{5}]

\frac{d [O_{2}]}{d t} = k_{3} [N_{2} O_{5}]

.
What is the relation between

k_{1}, k_{2}

and

k_{3}

Q. For a reaction,

N_{2} O_{5} ⟶ 2 N O_{2} + \frac{1}{2} O_{2}

Given:

\frac{- d [N_{2} O_{5}]}{d t} = k_{1} [N_{2} O_{5}]

\frac{d [N O_{2}]}{d t} = k_{2} [N_{2} O_{5}]

\frac{d [O_{2}]}{d t} = k_{3} [N_{2} O_{5}]

The relation between

k_{1}, k_{2}

and

k_{3}

are:

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For the reaction 2N2O5(g)→4NO2(g)+O2(g) −Δ[N2O5]Δt=k1[N2O5],+[NO2]Δt=k2[N2O5],+Δ[O2]Δt=k3[N2O5] Thus,

The correct option is B 2k1=k2=4k3 For the reaction 2N2O5(g)→4NO2(g)+O2(g) Rate of the reaction: −12d[N2O5]dt=k12N2O5] (i) 14d[NO2]dt=k24[N2O5] (ii) d[O2]dt=k3[N2O5] (iii) all are rate of reaction thus i=ii=iii 2k1=k2=4k3

For the reaction
$2 N_{2} O_{5} (g) \to 4 N O_{2} (g) + O_{2} (g)$
$- \frac{Δ [N_{2} O_{5}]}{Δ t} = k_{1} [N_{2} O_{5}], + \frac{[N O_{2}]}{Δ t} = k_{2} [N_{2} O_{5}], + \frac{Δ [O_{2}]}{Δ t} = k_{3} [N_{2} O_{5}]$ Thus,