Q. 84.2( 41 Votes )

# The rate of the chemical reaction doubles for an increase of 10K in absolute temperature from 298K. Calculate E_{a}.

Answer :

Given-

Initial temperature, T_{1} = 298 K

Final temperature, T_{2} = 298 K + 10 K = 308 K

Knowing that the rate constant of a chemical reaction normally increases with increase in temperature, we assume that,

Initial value of rate constant, k_{1} = k

Final value of rate constant, k_{2} = 2k

Using Arrhenius equation,

→ Equation 1

where, R = 8.314 J K^{-1} mol^{-1} (gas constant).

Substituting all the values in equation 1, we get,

log 2 =

E_{a} =

E_{a} =

E_{a} = 52897 J mol^{-1}

E_{a} = 52.897 kJ mol^{-1}

__The energy of activation, E _{a} is 52.897 kJ mol^{-1}__

Rate this question :

What happens to most probable kinetic energy and the energy of activation with increase in temperature?

Chemistry - ExemplarOxygen is available in plenty in air yet fuels do not burn by themselves at room temperature. Explain.

Chemistry - ExemplarWhy does the rate of a reaction increase with rise in temperature?

Chemistry - ExemplarWhich of the following graphs represents exothermic reaction?

(a)

(b)

(c)

Chemistry - Exemplar