Question

On the top of a certain mountain where atmospheric pressure is 535 mm Hg, pure water shows a boiling point of 360 K and it takes $3\times {10}^2$ min to boil at the top of the mountain as against 3 min taken to boil an egg at 370 K. The ratio of rate constants $\frac{K_{370}}{K_{360}}$ is ?

• A. $10$
• B. $1$
• C. $100$
• D. $\frac{1}{10}$

Answer 1

The correct option is C

$100$

From the problem it is evident that
Rate of boiling egg $\propto \frac{1}{times}=\frac{K}{t}$
or $K=t\times$ rate of boiling

Hence $K_{370}=3\times {10}^2\times rateofboiling$
$K_{360}=3\times$ rate of boiling
$\therefore \frac{K_{370}}{K_{360}}=\frac{3\times {10}^2}{3}={10}^{2}or100$