Question

Cyclopropane rearranges to form propene:

$△\to C{H}_2-CH=C{H}_2$

This follows first-order kinetics. The rate constant is $2.714\times {10}^{-3}se{c}^{-1}$. The initial concentration of cyclopropane is $0.29M$. What will be the concentration of cyclopropane after $100sec$?

• A. $0.035M$
• B. $0.22M$
• C. $0.145M$
• D. $0.0018M$

Answer 1

The correct option is B $0.22M$

$k=\frac{2.303}{t}\log \frac{a}{(a-x)}$

$(a-x)$ is the concentration left after $100sec$.

$2.7\times {10}^{-3}=\frac{2.303}{100}\log \frac{0.29}{(a-x)}$

$\Rightarrow \frac{0.27}{2.303}=\log \frac{0.29}{(a-x)}\Rightarrow 0.117=\log \frac{0.29}{(a-x)}$

$\Rightarrow (a-x)=0.22M$.

Hence, the correct answer is option $\text{B}$.