Home
Class 12
CHEMISTRY
The activation energy for the reaction -...

The activation energy for the reaction -
`H_(2)O_(2)toH_(2)O+(1)/(2)O_(2)`
is `18 K cal//mol` at 300 K. calculate the fraction of molecules of reactonts having energy equal to or greater than activation energy ?
Anti log `(-13.02)=9.36xx10^(-14)`

A

`9.36xx10^(-14)`

B

`1.2xx10^(-12)`

C

`4.2xx10^(-16)`

D

`5.2xx10^(-15)`

Text Solution

Verified by Experts

The correct Answer is:
A

`e^((Eq)/(RT))=e(18000)/(2xx300)=e^(-30)`
`=9.36xx10^(-14)`
Promotional Banner

Topper's Solved these Questions

  • CHEMISTRY AT A GLANCE

    ALLEN|Exercise INORGANIC CHEMISTRY|300 Videos
  • CHEMISTRY AT A GLANCE

    ALLEN|Exercise ORGANIC CHEMISTRY|483 Videos
  • Chemical Equilibrium

    ALLEN|Exercise All Questions|24 Videos
  • ELECTROCHEMISTRY

    ALLEN|Exercise EXERCISE -05 [B]|38 Videos

Similar Questions

Explore conceptually related problems

The activation energy for the reaction : ltbr. 2Hl(g) rarr H_(2)(g)+I_(2)(g) is 209.5 kJ mol^(-1) at 581K . Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy ?

The activation energy for the reaction, 2Hi(g) to H_(2)(g) + I_(2)(g) is 209.5 kJ moli^(-1) at 581 K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy.

The activation energy for the reaction: 2AB rarr A_(2)+B_(2)(g) is 159.7 kJ mol^(-1) at 500 K . Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy. (Given: 2.3 xx 8.314 J K^(-1) mol^(-1) xx 500 K = 9561.1 J mol^(-1) )

The activation energy for the decomposition of hydrogen iodide at 581 K is 209.5 kJ mol^(–1) . Calculate the fraction of molecules having energy equal to activation energy.

Assertion : For a chemical reaction with rise in temperature by 10^(@) the rate constant is nearly doubled. Reason : At t + 10, the fraction of molecules having energy equal to or greater than activation energy gets doubled.

The standard Gibbs energy change for the reaction 2SO_2(g) +O_2(g) to 2SO_3 (g) is -142 kJ mol^(-1) at 300 K. Calculate equilibrium constant.

The slope of the line graph of log k versus 1//T for the reaction N_(2)O_(5) rarr2NO_(2) + 1//2O_(2) is -5000 .Calculate the energy of activation of the reaction (in kJ K^(-1) mol^(-1) ).