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

`E_a=1124.83 J/mol`

Step-by-step 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(K_2)/(K_1)=(E_a)/(R)\left ( (1)/(T_1)-(1)/(T_2) \right )`

Now by putting the values we can get

`\ln(0.0815)/(0.0796)=(E_a)/(8.314)\left ( (1)/(1010)-(1)/(1220) \right )`

`0.023=0.00017* (E_a)/(8.314)`

`E_a=0.023* (8.314)/(0.00017)`

`E_a=1124.83 J/mol`

Therefore the activation energy will be 1124.83 J/mol