Question

Understanding the high-temperature formation and breakdown of the nitrogen oxides is essential for controlling the pollutants generated by car engines. The second-order reaction for the breakdown of nitric oxide to its elements has rate constants of 0.0796 L/mol-s at 737°C and 0.0815 L/mol-s at 947°C. What is the activation energy of this reaction? Give your answer in scientific notation.

Answer 1

Answer:

$1.143*10^{3} J/mol$

Explanation:

Using the Arrhenius equation for the give problem:

$k = A*exp^{\frac{-E_{a} }{RT} }$

Taking the natural log (ln) of both sides, we have:

$ln k = ln A - \frac{E_{a} }{RT}$

In the given problem, we have two rate constants at two different temperatures. Thus:

$ln k_{1} = ln A - \frac{E_{a} }{RT_{1} }$            (1)

$ln k_{2} = ln A - \frac{E_{a} }{RT_{2} }$          (2)

Subtracting equation (1) from equation (2), we have:

$ln k_{2} - ln k_{1} = \frac{E_{a} }{RT_{1} } - \frac{E_{a} }{RT_{2} }$      (3)

$k_{1} = 0.0796 L/mol-s$;   $T_{1} = 737°C = 737+273 = 1010 K$

$k_{2} = 0.0815 L/mol-s$;   $T_{2} = 947°C = 737+273 = 1220 K$

Therefore, equation (3) becomes:

$ln 0.0815 - ln 0.0796 = E_{a}[\frac{1}{8.314*1010} - \frac{1}{8.314*1220}]$

-2.507 - (2.531) = Ea*[0.00012 - 0.000099]

Ea = 0.024/0.000021 = 1142.86 J/mol

The activation energy of the reaction in scientific notation is $1.143*10^{3} J/mol$