Question

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

• A. $ 1.3\times {{10}^{-5}}M{{s}^{-1}} $
• B. $ 0.5\times {{10}^{-4}}M{{s}^{-1}} $
• C. $ 7.6\times {{10}^{-4}}M{{s}^{-1}} $
• D. $ 2\times {{10}^{-3}}M{{s}^{-1}} $
• E. $ 2.5\times {{10}^{-5}}M{{s}^{-1}} $

Answer 1

Answer: (A)

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