JEE Main 2014 · 11 Apr · Shift 1 · Chemistry Q35

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ChemistrySingle correct+4 / -1Question 35 of 90

For the reaction, $\ce{2N2O5 -> 4NO2 + O2}$ , the rate equation can be expressed in two ways $-\dfrac{d[\ce{N2O5}]}{dt}=k[\ce{N2O5}]$ and $+\dfrac{d[\ce{NO2}]}{dt}=k'[\ce{N2O5}]$

k and k′ are related as :

k=k'

2k=k'

k=2k'

k=4k'

Answer and worked solution

Official answer: B

Four $\ce{NO2}$ form for every two $\ce{N2O5}$ used:
$\frac{d[\ce{NO2}]}{dt}=2\left(-\frac{d[\ce{N2O5}]}{dt}\right)=2k[\ce{N2O5}].$
So $k'=2k$ .

Topics:Rate of reaction and rate law

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JEE Main 2014 · 11 Apr · Shift 1, Chemistry question 35