When liquid benzene is oxidised at constant pressure at 300 k, The change in enthalpy is -3728 kj.What is the change in internal energy at the sam temperature? (R=8.314xx10-3 kj K-1 mol-1)

Latent heat of vaporisation of a liquid at 500 K and 1 atm pressure is 10 kcal//mol . What will be the change in internal energy (DeltaU) of 3 mol of liquid at the same temperature ?

If Delta H and Delta U are the changes in the enthalpy and internal energy when a liquid is converted into its vapours at temperature T, then :

Calculate the number of moles of hydrogen present in 500 cm^(3) of a gas under a pressure of 101.3 kPA at a temperature of 300K . ( R=8.314 J K^(-1) mol^(-1)) .

Calculate the number of moles of hydrogen present in 500 cm^(3) of a gas under a pressure of 101.3 kPA at a temperature of 300K . ( R=8.314 J K^(-1) mol^(-1)) .

The rate constant of a reaction is doubled when the temperature increased from 400K to 410K. Calculate the activation energy (Ea) [R = 8.314 JK^(-1) " mol"^(-1)] .

Rate constant of a reaction is k = 3.14 xx 10^(-4) mol L^(-1)s^(-1) . What is the order of the reaction.

In general it is observed that the rate of a chemical reaction doubles with every 10 degree rise in temperature. If the generalization holds good for the reaction in the temperature range 295 K to 305 K, what would be the value of activation energy for this reaction ? [R = 8.314 mol^(-1)JK^(-1)]