Question

Question 24 of 30 > A certain reaction has an enthalpy of AH = 34 kJ and an activation energy of Ea = 66 kJ. What is the activation energy of the reverse reaction? Ea(reverse) = kJ

Answer 1

For the 24^th question, let us solve it with our basic knowledge from the chemical kinetics topic

Threshold energy intermediate activated Complex Er Ea ER ΔΗ 66aus Ep Progress of reaction ER= Energy of reactant Ep = Energy of product

The above graph tell us the answer for the given question; let us first define activation energy.

Activation energy is the difference in energy of intermediate activated complex and the average energy of reactants.

Now , in the reverse reaction products will be treated as reactants and reactants will be treated as products , then the reverse activation energy will be difference in energy of intermediate activated complex and the average energy of products(i.e, they behave as reactants in reverse reaction)

Ea(rev)=$\delta$H+Ea

= 34+66 KJ

= 100 KJ

For the 23^th Question,

From the 24^th question diagram ,we can mention some labels like energy of reactant , energy of product ,change in enthalphy , activation energy .

rate limiting Tịansition state non-limiting transition state Activation Energy Energy R. + HX Z intermediates RX + H starting material Enthalpy change RH + X products. Reaction coordinate

let us take a example

A--(step 1)->B-(step 2)-->C

Ei B E CA С.

If E₁ > E3 then step(i) is slow and as we know slowest step is rate determinig step

Here E₁ < E3then step (ii) is slowest and rate determinig step

A* E. Ei Ez ha A B С

this step labels you rate limiting transition state for the respective slowest step and non-limiting transition for another step