Question

In Arrhenius equation for a certain reaction, the value of A (Arrhenius factor) and $E_a$ (activation energy) are $4\times {10}^{13}{s}^{-1}\text{and}98.6kJmo{l}^{-1}$ respectively. At what temperature, the reaction will have specific rate constant of $1.1\times {10}^{-3}{s}^{-1}$?
Take
$log\left(\frac{1.1}{4}\right)=-0.56$

• A. $79K$
• B. $167K$
• C. $273K$
• D. $311K$

Answer 1

The correct option is D $311K$

According to Arrhenius the relationship between temperature and rate constant is given by,
$k=A{e}^{-E_a/RT}$
where,
k : Rate constant​
A : Arrhenius factor or frequency factor or pre-exponential factor​
$E_a$ : Activation energy (in J/mol)​
R : Gas constant​
T : Temperature of reaction (in Kelvin(K))

$k=A{e}^{-E_a/RT}$
Taking ln on both side

$lnk=lnA-\frac{E_a}{RT}$

$2.303logk=2.303logA-\frac{E_a}{RT}$
$2.303log(1.1\times {10}^{-3})=2.303log(4\times {10}^{13})-\frac{98.6\times {10}^3}{8.314\times T}$
$2.303\left(log\frac{1.1\times {10}^{-3}}{4\times {10}^{13}}\right)=-\frac{98.6\times {10}^3}{8.314\times T}$
$2.303(log\left(\frac{1.1}{4}\right)-16)=-\frac{98.6\times {10}^3}{8.314\times T}$
$-2.303\times 16.56=-\frac{98.6\times {10}^3}{8.314\times T}$

$T=\frac{98.6\times {10}^3}{8.314\times 2.303\times 16.56}$

$=310.96K$