Question

The reaction between nitrogen dioxide and carbon monoxide is NO2(g)+CO(g)→NO(g)+CO2(g)NO2(g)+CO(g)→NO(g)+CO2(g) The rate constant at 701 KK is measured as 2.57 M−1⋅s−1M−1⋅s−1 and that at 895 KK is measured as 567 M−1⋅s−1M−1⋅s−1. The activation energy is 1.5×102 1.5×102 kJ/molkJ/mol. Predict the rate constant at 525 KK . Express your answer in liters per mole-second to three significant figures.

Answer 1

Answer:

$k_2=2.55\ {Ms}^{-1}$

Explanation:

Using the expression,

$\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )$

Where,

$k_1\ is\ the\ rate\ constant\ at\ T_1$

$k_2\ is\ the\ rate\ constant\ at\ T_2$

$E_a$ is the activation energy

R is Gas constant having value = 8.314 J / K mol  

$k_2=?$

$k_1=2.57\ {Ms}^{-1}$

$T_1=701\ K$  

$T_2=525\ K$  

$E_a=1.5\times 10^2\ kJ/mol$

So,  

$\ln \:\frac{2.57}{k_2}\:=-\frac{1.5\times \:10^2}{8.314}\times \left(\frac{1}{701}-\frac{1}{525}\right)\:\:$

$k_2=\frac{2.57}{e^{\frac{352}{40796.798}}}=2.55\ {Ms}^{-1}$