Chapter 13
Q. 13.8
using the Two-Point Form of the arrhenius Equation
The reaction between nitrogen dioxide and carbon monoxide is:
NO_{2}(g) + CO(g) → NO(g) + CO_{2}(g)
The rate constant at 701 K is measured as 2.57 M^{-1} . s^{-1}, and that at 895 K is measured as 567 M^{-1} . s^{-1}. Find the activation energy for the reaction in kJ/mol.
| SORT You are given the rate constant of a reaction at two different tem-peratures and asked to find the activation energy. | GIVEN T_{1} = 701 K, k_{1} = 2.57 M^{-1} . s^{-1} T_{2} = 895 K, k_{2} = 567 M^{-1} . s^{-1} FIND E_{a} |
| STRATEGIZE Use the two-point form of the Arrhenius equation, which relates the activation energy to the given information and R (a constant). | EQUATION ln \frac{k_{2}}{k_{1}}= \frac{E_{a}}{R} \left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right) |
| SOLVE Substitute the two rate constants and the two temperatures into the equation. Solve the equation for E_{a}, the activation energy, and convert to kJ/mol. |
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Step-by-Step
Verified Solution
ln\frac{ 567\cancel{ M^{-1} . s^{-1}}}{2.57\cancel{ M^{-1} . s^{-1}}} = \frac{E_{a}}{R} \left(\frac{1}{701K}-\frac{1}{895K}\right)
5.40 = \frac{E_{a}}{R} \left(\frac{3.\underline{0}9\times10^{-4}}{K}\right)
E_{a} = 5.40 \left(\frac{K}{3.\underline{0}9\times10^{-4}}\right)R
= 5.40 \left(\frac{\cancel{K}}{3.\underline{0}9\times10^{-4}}\right) 8.314 \frac{J}{mol . \cancel{K}}
= 1.5 × 10^{5} J/mol
= 1.5 × 10² kJ/mol
CHECK The magnitude of the answer is reasonable. Activation energies for most reactions range from tens to hundreds of kilojoules per mole.