Question

If $50\%$ of a reaction occurs in $100$ second and $75\%$ of the reaction occurs in $200$ second, the order of this reaction is :

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (B)

For a first order reaction $k=\frac{2.303}{t}\log \frac{a}{a-x}$

At $100s:k=\frac{2.303}{100}\log \frac{100}{100-50}$

$k=\frac{2.303}{100}\log 2=\frac{2.303\times 0.3010}{100}=0.00693mi{n}^{-1}$

At $200s:k=\frac{2.303}{200}\log \frac{100}{100-75}$

$k=\frac{2.303}{200}\log 4=\frac{2.303\times 0.6020}{200}=0.00693mi{n}^{-1}$

As rate constant is same at $100s$ and $200s$, therefore, order of reaction is 1 .