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Byju's Answer

Standard XII

Chemistry

First Order Reaction

The reaction,...

Question

The reaction, $2 N_{2} O_{5} (g) ⟶ 4 N O_{2} (g) + O_{2} (g)$ , is first order with respect to $N_{2} O_{5}$ . Which of the following graph, would yield a straight line?

log (p_{N_{2} O_{5}})

versus time with -ve slope

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{(p_{N_{2} O_{5}})}^{- 1}

versus time

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

versus time

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log (p_{N_{2} O_{5}})

versus time with +ve slope

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Solution

The correct option is A $log (p_{N_{2} O_{5}})$ versus time with -ve slope
$V = k . p_{N_{2} O_{5}}$
So, $log (\frac{p_{N_{2} O_{5}}}{{p^{0}}_{N_{2} O_{5}}}) = - k t ∴ log (p_{N_{2} O_{5}}) = - k t + log {p^{0}}_{N_{2} O_{5}} log (p_{N_{2} O_{5}}) = - k t + C$
So stoke is $- k$ for $log (p_{N_{2} O_{5}})$ as $t .$

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Similar questions

Q. Calculate the average rate of decomposition of

N_{2} O_{5}

by the reaction,

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

.
During the time interval from

t = 600 s

t = 1200 s

using the following data:

\begin{matrix} T i m e & [N_{2} O_{5}] 600 s & 1.24 \times 10^{- 2} M 1200 s & 0.93 \times 10^{- 2} M \end{matrix}

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 ?

Q. The rate constant for the reaction,

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

, is

2.3 \times 10^{- 2} {s e c}^{- 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}

corresponds to concentration of

N_{2} O_{5}

initially and time

t

respectively?

Q. The reaction

N_{2} O_{5} (I n {C C l}_{4}) \to 2 {N O}_{2} + 1 / 2 O_{2} (g)

is first order in

N_{2} O_{5}

with rate constant

6.2 \times 10^{- 4} S^{- 1}

. What is the value of rate of reaction when

[N_{2} O_{5}] = 1.25 m o l e L^{- 1}

Q. Calculate

Δ H / k J

for the following reaction using the listed standard enthalpy of reaction data:

2 N_{2} (g) + 5 O_{2} (g) ⟶ 2 N_{2} O_{5} (s)

N_{2} (g) + 3 O_{2} (g) + H_{2} (g) ⟶ 2 H N O_{3} (a q) Δ H / k J = - 414.0

N_{2} O_{5} (s) + H_{2} O (I) ⟶ 2 H N O_{3} (a q) Δ H / k J = - 86.0

2 H_{2} (g) + O_{2} (g) ⟶ 2 H_{2} O (I) Δ H / k J = - 571.6

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The reaction, 2N2O5(g)⟶4NO2(g)+O2(g), is first order with respect to N2O5. Which of the following graph, would yield a straight line?

The correct option is A log(pN2O5) versus time with -ve slopeV=k.pN2O5 So, log(pN2O5p0N2O5)=−kt∴log(pN2O5)=−kt+logp0N2O5log(pN2O5)=−kt+C So stoke is −k for log(pN2O5) as t.

The reaction, $2 N_{2} O_{5} (g) ⟶ 4 N O_{2} (g) + O_{2} (g)$ , is first order with respect to $N_{2} O_{5}$ . Which of the following graph, would yield a straight line?