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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 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.

Respuesta :

Answer:

[tex]k_2=2.55\ {Ms}^{-1}[/tex]

Explanation:

Using the expression,

[tex]\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )[/tex]

Where,

[tex]k_1\ is\ the\ rate\ constant\ at\ T_1[/tex]

[tex]k_2\ is\ the\ rate\ constant\ at\ T_2[/tex]

[tex]E_a[/tex] is the activation energy

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

[tex]k_2=?[/tex]

[tex]k_1=2.57\ {Ms}^{-1}[/tex]

[tex]T_1=701\ K[/tex]  

[tex]T_2=525\ K[/tex]  

[tex]E_a=1.5\times 10^2\ kJ/mol[/tex]

So,  

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

[tex]k_2=\frac{2.57}{e^{\frac{352}{40796.798}}}=2.55\ {Ms}^{-1}[/tex]

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