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

`1.143*10^(3) J/mol`

Step-by-step explanation:

Using the Arrhenius equation for the give problem:

`k = A*exp^{(-E_(a) )/(RT) }`

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

`ln k = ln A - (E_(a) )/(RT)`

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

`ln k_(1) = ln A - (E_(a) )/(RT_(1) )` (1)

`ln k_(2) = ln A - (E_(a) )/(RT_(2) )` (2)

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

`ln k_(2) - ln k_(1) = (E_(a) )/(RT_(1) ) - (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)[(1)/(8.314*1010) - (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`