For the second-order reaction NO( g) + O 3( g) → NO 2( g) + O 2( g), the rate constant has been measured to be 1.08 × 10 7 M –1 s –1 at 298 K and the activation energy has been measured to be 11.4 kJ/mol over the temperature range 195 K to 304 K. What is the rate constant at 207 K? ( R = 8.3145 J K –1 mol –1)

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Alleei Alleei · Answer 1 · 2020-04-22T08:25:19+02:00

Answer : The rate constant at 207 K is, [tex]4.49\times 10^6M^{-1}s^{-1}[/tex]

Explanation :

According to the Arrhenius equation,

[tex]K=A\times e^{\frac{-Ea}{RT}}[/tex]

or,

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

where,

[tex]K_1[/tex] = rate constant at [tex]298K[/tex] = [tex]1.08\times 10^7M^{-1}s^{-1}[/tex]

[tex]K_2[/tex] = rate constant at [tex]207K[/tex] = ?

[tex]Ea[/tex] = activation energy for the reaction = [tex]11.4kJ/mol=11400J/mol[/tex]

R = gas constant = 8.314 J/mole.K

[tex]T_1[/tex] = initial temperature = 298 K

[tex]T_2[/tex] = final temperature = 207 K

Now put all the given values in this formula, we get:

[tex]\log (\frac{K_2}{1.08\times 10^7M^{-1}s^{-1}})=\frac{11400J/mol}{2.303\times 8.314J/mole.K}[\frac{1}{298K}-\frac{1}{207K}][/tex]

[tex]K_2=4.49\times 106M^{-1}s^{-1}[/tex]

Therefore, the rate constant at 207 K is, [tex]4.49\times 10^6M^{-1}s^{-1}[/tex]