Home
Class 12
CHEMISTRY
Find out the percentage of the reactant ...

Find out the percentage of the reactant molecules crosisng over the energy barrier at `325 K`.
Given: `Delta H_(325 K) = 0.12 kcal`,
`E_(a(b)) = 0.02 kcal`

Text Solution

Verified by Experts

The correct Answer is:
`80.65%;`
Promotional Banner

Topper's Solved these Questions

  • CHEMICAL KINETICS

    P BAHADUR|Exercise Exercise 1|2 Videos
  • CHEMICAL KINETICS

    P BAHADUR|Exercise Exercise 3A|2 Videos
  • DILUTE SOLUTION AND COLLIGATIVE PROPERTIES

    P BAHADUR|Exercise Exercise 9 Advanced Numerical|12 Videos

Similar Questions

Explore conceptually related problems

What percentage of reactant molecules will crossover the energe barrier t 325 K? Heat of reaction is 0.12 kcal and activation energy of backward reaction is 0.02 kcal.

Given that the enthalpy change of a reaction at 325 K is 0.12 kcal and energy of activation for the backward reaction is 0.02 kcal, calculate the percentage of the reactant molecules, crossing over the energy barrier.

Calculate the value of equilibrium constant for the reaction: A_((g))+B_((g))hArrC_((g))+D_((g))+E_((g)) at 300 K and constant pressure. All the reactants and products obey ideal gas nature. Given DeltaU^(@)=-90.0 kcal , DeltaS^(@)=100 calK^(-1) .

Find the values of k if area of triangle is 4 sq. units and vertices are: (k,0),(0,2)(4,0)

Calculate the lattice energy from the following data (given 1 eV = 23.0 kcal mol^(-1) ) i. Delta_(f) H^(ɵ) (KI) = -78.0 kcal mol^(-1) ii. IE_(1) of K = 4.0 eV iii. Delta_("diss")H^(ɵ)(I_(2)) = 28.0 kcal mol^(-1) iv. Delta_("sub")H^(ɵ)(K) = 20.0 kcal mol^(-1) ltbvrgt v. EA of I = -70.0 kcal mol^(-1) vi. Delta_("sub")H^(ɵ) of I_(2) = 14.0 kcal mol^(-1)

When 12.0 g of carbon reacted with limited quantity of oxygen, 57.5 kcal of heat was produced, calculate the number of moles of CO produced (Delta_(f)H(CO_(2))=-94.5 cal, Delta_(f)H(CO)= -21.41kcal.

From the following data at 25^(@)C , calculate the bond energy of O – H bond: (i) H_(2)(g) to 2H(g), Delta H_(1) = 104.2 kcal (ii) O_(2)(g) to 2O(g), DeltaH_(2) = 118.4 kcal (iii) H_(2)(g) + 1/2 O_(2)(g) to H_(2)O(g), Delta H_(3) = -57.8 k cal