Question

For a chemical reaction $ A\to B, $ the rate of the reaction is $ 2\times {{10}^{-3}} $ mol $ d{{m}^{-3}}{{s}^{-1}}, $ when the initial concentration is 0.05 mol $ d{{m}^{-3}} $ . The rate of the same reaction is $ 1.6\times {{10}^{-2}}mol\text{ }d{{m}^{-3}}{{s}^{-1}} $ when the initial concentration is $ 0.1\text{ }mol\text{ }d{{m}^{-3}} $ . The order of the reaction is

• A. 2
• B. 0
• C. 3
• D. 1

Answer 1

Answer: (C)

$ A\xrightarrow{{}}B $ Rate $ =k{{[A]}^{n}} $ $ {{(Rate)}_{1}}=k{{(0.05)}^{n}}=2\times {{10}^{-3}} $ $ {{(Rate)}_{2}}=k{{(0.1)}^{n}}=1.6\times {{10}^{-2}} $ Dividing the Eq. (ii) by Eq. (i) $ \frac{{{(Rate)}_{2}}}{{{(Rate)}_{1}}}=\frac{k{{(0.1)}^{n}}}{k{{(0.05)}^{n}}}=\frac{1.6\times {{10}^{-2}}}{2\times {{10}^{-3}}} $ $ {{(2)}^{n}}=8 $ or $ {{(2)}^{n}}={{2}^{3}} $ $ \therefore $ $ n=3 $