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Question
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`
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