Question

The rate of a first order reaction gets doubled when the temperature is doubled from $T_{1}to2T_1$ (in K). $E_a$ is nearly constant activation energy at 1386 kcal. What is the value of $T_1$?
(given R=2 $cal{K}^{-1}mo{l}^{-1}$,ln2=0.693)

• A. 250 K
• B. 500 K
• C. 1000 K
• D. 750 K

Answer 1

The correct option is B 500 K

According to the Arrhenius equation,
$k=A{e}^{\frac{-E_a}{RT}}$
Taking logarithm we will get,
$ln{k}_1=lnA-\frac{E_a}{R{T}_1}$.....(1)
$ln{k}_2=lnA-\frac{E_a}{R(2T{)}_1}$.......(2)
Subtracting equation (1) from (2) we get
$ln\frac{k_2}{k_1}=\frac{E_a}{R}(\frac{1}{T_1}-\frac{1}{2T_1})\Rightarrow ln2=\frac{1386}{2}(\frac{1}{T_1}-\frac{1}{2T_1})\Rightarrow T_1=\frac{1386}{0.693\times 4}=500K$