N2O5 g - 2NO2 g -1-2O2 g In the above first order reaction the initial concentration of N2O5 is 2-40 -10-2 mol L-1 at 318 K- The concentration of N2O5 after 1 hour was 1-60- 10-2mol L-1 - The rate constant of the reaction at 318 k is x-10-3 min -1 The value of x Nearest integer is - Given- log 3-0-447 - log 5-0-699 -

For the first order reaction nbsp; kt= ln [ R ]_0 [ R ] [R_0] is the initial concentration [R] is the final concentration at time t k× 60=ln ( 2.4× 10^ 2 ...

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Byju's Answer

Standard XII

Chemistry

Methods for Ester Synthesis

N2O5 g → 2NO2...

Question

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

In the above first order reaction the initial concentration of $N_{2} O_{5}$ is 2.40 $\times 10^{- 2} m o l L^{- 1}$ at 318 K. The concentration of $N_{2} O_{5}$ after 1 hour was $1.60 \times 10^{- 2} m o l L^{- 1}$ . The rate constant of the reaction at 318 k is $x \times 10^{- 3} m i n^{- 1}$ The value of x (Nearest integer) is

$[G i v e n : l o g 3 = 0.447, l o g 5 = 0.699]$

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Solution

For the first order reaction

$k t = ln \frac{{[R]}_{0}}{[R]}$
$[R_{0}]$ is the initial concentration
$[R]$ is the final concentration at time t

$k \times 60 = ln \frac{(2.4 \times 10^{- 2})}{(1.6 \times 10^{- 2})}$

$k = 2.303 \times (l o g 3 - l o g 2)$

$k = 2.303 \times (0.477 - 0.301)$

$k = 6.7 \times 10^{- 3} m i n^{- 1}$
$k \approx 7 \times 10^{- 3} m i n^{- 1} x = 7$

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Similar questions

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 initial concentration of

N_{2} O_{5}

in the first order reaction

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

was 1

1.24 \times 10^{- 2} m o l L^{- 1}

at 318 K. The concentration of

N_{2} O_{5}

after 60 minutes was

{0.20 \times 10}^{- 2} m o l 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. The reaction;

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

is of first order for

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 L^{- 1}

Q. For the reaction,

2 N_{2} O_{5} ⇌ 4 N O_{2} + O_{2}

rate and rate constant are

2.40 \times 10^{- 5}

mol

L^{- 1} s^{- 1}

and

3 \times 10^{- 5} s^{- 1}

respectively, then calculate the concentration of

N_{2} O_{5}

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For the first order reaction kt=ln[R]0[R] [R0] is the initial concentration [R] is the final concentration at time t k×60=ln(2.4×10−2)(1.6×10−2) k=2.303×(log3−log2) k=2.303×(0.477−0.301) k=6.7×10−3min−1 k≈7×10−3min−1x=7