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 constan

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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 constan

t 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 .

Chemistry

1 answer:

eduard [2.7K] · Answer 1 · 2026-03-03T02:55:25+03:00

Response: The rate constant at 525 K is, $0.0606M^{-1}s^{-1}$

Rationale:

Based on the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

or,

$\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_1$ = rate constant when $701K$ = $2.57M^{-1}s^{-1}$

$K_2$ = rate constant when $525K$ =?

$Ea$ = activation energy for the process = $1.5\times 10^2kJ/mol=1.5\times 10^5J/mol$

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = 701 K

$T_2$ = final temperature = 525 K

Substituting the provided values into this formula yields:

$\log (\frac{K_2}{2.57M^{-1}s^{-1}})=\frac{1.5\times 10^5J/mol}{2.303\times 8.314J/mole.K}[\frac{1}{701K}-\frac{1}{525K}]$

$K_2=0.0606M^{-1}s^{-1}$

Thus, the rate constant at 525 K is, $0.0606M^{-1}s^{-1}$

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 constan

5060 has three significant figures: Below is the clarification

s < p < d < f