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Following Data Are Obtained for Reaction :N2o5 → 2no2 + 1/2o2 Show that It Follows First Order Reaction. Calculate the Half-life - Chemistry

Following data are obtained for reaction :

N2O5 → 2NO2 + 1/2O2

t/s 0 300 600
[N2O5]/mol L–1 1.6 × 10-2 0.8 × 10–2 0.4 × 10–2

1) Show that it follows first order reaction.

2) Calculate the half-life.

(Given log 2 = 0.3010, log 4 = 0.6021)

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Solution

1) For 1st order reaction the integral rate law is :

kt = `ln  a_0/a_t`

Given

a0 = 1.6×10−2 mol L−1

For t = 300 s, at = 0.8×10−2 mol L−1

For t = 600 s, at = 0.4×10−2 mol L−1

Using first set of data in the rate law,

`k xx 300 = ln (1.6 xx 10^(-2))/(0.8xx 10^(-2))`

k = `0.00231 s^(-1)`

Using second set of data in the rate law, 

`k xx 600 = ln  (1.6 xx 10^(-2))/(0.4 xx 10^(-2))`

k = 0.00231 s-1

The value of k is consistent, therefore it follows first order reaction.

2) The half life of first order reaction is given by the following equation:

`t_(1/2) = (ln 2)/k = 2.303 xx (log 2)/k`

`:. t_(1/2) = 2.303 xx (log 2)/0.00231 = 300.08 s`

Concept: Integrated Rate Equations - First Order Reactions
  Is there an error in this question or solution?
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