English

The decomposition of N2O5 in CCl4 solution follows the first order rate law. The concentration of N2O5 measured at different time intervals are given below. Time in seconds (t) 0 80 160 410 600 1130

Advertisements
Advertisements

Question

The decomposition of N2O5 in CCl4 solution follows the first order rate law. The concentration of N2O5 measured at different time intervals are given below.

Time in seconds (t) 0 80 160 410 600 1130 1720
[N2O5] (mol/L) 5.5 5.0 4.8 4.0 3.4 2.4 1.6

Calculate its rate constant at t = 410 s and t = 1130 s. What do these results show?

Numerical
Advertisements

Solution

Given:

The decomposition of N2O5 follows first-order kinetics

We are to calculate the rate constant at two time intervals:

t = 410 s

t = 1130 s

By using first-order rate law

`k = 2.303/t log ([A]_0/([A]))`

Let, [A]0​ = 5.5 mol/L at t = 0

1. At t = 410 s, [A] = 4.0 mol/L

`k_410 = 2.303/410 log (5.5/4.0)`

= `2.303/410 log (1.375)`

= `(2.303 xx 0.1379)/410`

= `0.3177/410`    ...(log 1.375 = 0.1379)

k410 = 7.75 × 10−4 s1

2. At t = 1130 s, [A] = 2.4 mol/L

`k_1130 = 2.303/1130 log (5.5/2.4)`

= `2.303/1130 log (2.2917)`

= `(2.303 xx 0.3603)/1130`    ...(log 2.2917 = 0.3603)

= `0.8299/1130`

k1130 = 7.34 × 10−4 s1

shaalaa.com
  Is there an error in this question or solution?
Chapter 3: Chemical Kinetics - REVIEW EXERCISES [Page 237]

APPEARS IN

Nootan Chemistry [English] Class 12 ISC
Chapter 3 Chemical Kinetics
REVIEW EXERCISES | Q 4.52 | Page 237
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×