Question

nderstanding 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:

$E_a=1124.83 J/mol$

Explanation:

Given that second order equation

K₁ = 0.0796 L/mol-s , T₁= 737⁰C

T₁ = 737 + 273 K = 1010 K

K₂ = 0.0815 L/mol-s , T₂=947°C

T₂=947+273 K= 1220 K

The activation energy given as follows

$\ln\dfrac{K_2}{K_1}=\dfrac{E_a}{R}\left ( \dfrac{1}{T_1}-\dfrac{1}{T_2} \right )$

Now by putting the values we can get

$\ln\dfrac{0.0815}{0.0796}=\dfrac{E_a}{8.314}\left ( \dfrac{1}{1010}-\dfrac{1}{1220} \right )$

$0.023=0.00017\times \dfrac{E_a}{8.314}$

$E_a=0.023\times \dfrac{8.314}{0.00017}$

$E_a=1124.83 J/mol$

Therefore the activation energy will be 1124.83 J/mol