Consider a reaction with an activation energy of 11186.5 J/mol run at 298.15 K and at 2226 K. How many times larger is the rate constant, k k , for the reaction at 2226 K than at 298 K

i think a, hope it .

explanation:

the answer is learned

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :