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Chemistry Question on Rate of a Chemical Reaction

For the reaction 2N2O5(g)→4NO2(g)+O2(g)2{{N}_{2}}{{O}_{5}}(g)\xrightarrow[{}]{{}}4N{{O}_{2}}(g)+{{O}_{2}}(g) if the concentration of NO2N{{O}_{2}} increases by 5.2×10−35.2\times {{10}^{-3}} M in 100 s then the rate of the reaction is:

A

1.3×10−5Ms−11.3\times {{10}^{-5}}M{{s}^{-1}}

B

0.5×10−4Ms−10.5\times {{10}^{-4}}M{{s}^{-1}}

C

7.6×10−4Ms−17.6\times {{10}^{-4}}M{{s}^{-1}}

D

2×10−3Ms−12\times {{10}^{-3}}M{{s}^{-1}}

Answer

1.3×10−5Ms−11.3\times {{10}^{-5}}M{{s}^{-1}}

Explanation

Solution

Rate of reaction =14d(NO2)dt=\frac{1}{4}\frac{d(N{{O}_{2}})}{dt}
= 5.2×10−34×100\frac{5.2\times {{10}^{-3}}}{4\times 100}
=1.3×10−5Ms−1=1.3\times {{10}^{-5}}M{{s}^{-1}}