Question

Calculate the rate constant at 310 K, when rate constant at $300$ K is $1.6\times {10}^5$.

• A. $2.363\times {10}^5$
• B. $2.4\times {10}^5$
• C. $2.450\times {10}^5$
• D. $3.123\times {10}^5$

Answer 1

The correct option is A $2.363\times {10}^5$

Now $T_1=300K$, $T_2={37}^{\circ }C=310K$

$E_a=30146Jmo{l}^{-1}$

log $\frac{k_{310}}{k_{300}}=\frac{30146Jmo{l}^{-1}}{2.303\times 8.314}\times \frac{310-300}{300\times 310}=0.6193$

$\therefore$ $\frac{k_{310}}{k_{300}}=antilog\left(0.1693\right)=1.477$

Higher the rate constant, faster is the reaction, i.e., lesser is the time taken, hence

$\frac{t_{310}}{t_{300}}=\frac{k_{300}}{k_{310}}=\frac{I}{1.477}$

$\therefore t_{310}=t_{300}\times \frac{1}{1.477}=\frac{48}{1.477}=32.5hr$

$\frac{k_{310}}{k_{300}}|=1.477$
$\therefore k_{310}=1.477\times k_{3}00=1.477\times 1.6\times {10}^5$
$=2.363\times {10}^5$