reaction has rate constant `1.73 xx10^(-3) min^(-1)` and `4.86xx10^(-3) min^(-1)` at 300 K and 330 K respectively. Calculate the energy of activation of this reaction. [Given: R = 8.314 JK^(-1)` mol^(-1)`

For a first order reaction, the frequency factor and rate constant at 600K are 1.6xx10^(13) s^(-1) and 2.05xx10^(-8)s^(-1) respectively. Find the energy of activation.

The half life of a first order reaction having rate constant k=1.7xx10^(-5) sec^(-1) is :

The half life of a first order reaction is 900 min at 820 K.Estimate its half life at 720 K if the energy of activation of the reaction is 250 kJ mol^(-1)

Calculate the half life of a first order reaction from their rate constants given below:- 2 min^(-1)

For a gas the Henry's law constant is 1.25xx10^(-3) mol dm^(-3) bar .^(-1) at 50^@C . Calculate the solubility of the given gas at 5 bar and 50^@C ,.

5 moles of helium expand isothermally and reversibly form a pressure 40 xx 10-5 Nm^(-2) to 4 xx 10^(-5) at 300K Calculate the work done, change in internal energy and heat absorbed during the expansion. (R-8.314Jk^(-1)mol^(-1))

For the reaction : 2A + B rarr A_2B the rate= k[A][B]^2 with k = 2.0 xx 10^(-6) mol^(-2) L^2 S^(-1) . Calculate the initial rate of the reaction when [A] = 0.1 mol L^(-1), [B] = 0.2 mol L^(-1) . Calculate the rate of reaction after [A] is reduced to 0.06 mol L^(-1)

K_p for the reaction. 2SO_(2(g)) O2_((g)) rarr 2SO_(3(g)) is 7.1 x 10^(24) at 298 K . Calculate DeltaG° for the reaction. R = 8.314 J K^(-1) mol^(-1) .