Question

The rate constant of the chemical reaction doubled for an increase of $10 \,K$ in absolute temperature from $295\, K$. Calculate $ E_{a} $ .

• A. $ 51.8\,kJ\,mol^{-1} $
• B. $ 82.1\,kJ\,mol^{-1} $
• C. $ 23.8\,kJ\,mol^{-1} $
• D. $ 62.1\,kJ\,mol^{-1} $

Answer 1

Answer: (A)

According to the Arrhenius equation.
$\log \frac{ K _{2}}{ K _{1}}=\frac{ E _{ a }}{2.303 R }\left[\frac{1}{ T _{1}}-\frac{1}{ T _{2}}\right]$
$\frac{ K _{2}}{ K _{1}}=2$;
$T _{1}=295 \,K ; T _{2}=305 \,K$
$R =8.314\, JK ^{-1}\, mol ^{-1}$
$\therefore \log 2=\frac{E_{a}}{2.303 \times\left(8.314\, JK ^{-1} mol ^{-1}\right)}\left[\frac{1}{295\, K }-\frac{1}{305 \,K }\right]$
or $0.3010=\frac{ E _{ a }}{8.314 \,JK ^{-1} mol ^{-1}} \times \frac{(10)}{295 \times 305}$
$E _{ a }=\frac{0.3010 \times 2.303 \times 8.314 \times 295 \times 305}{10}$
$=51860 \,J \,mol ^{-1}=51.86\, KJ \,mol ^{-1}$