For the process to occur under adiabatic conditions, the correct condition is:
(i) ∆T= 0 (ii) ∆p = 0
(iii) q = 0 (iv)  w = 0

 The enthalpies of all elements in their standard states are : ‘
(i) unity (ii) zero
(iii) < 0 (iv) different for each element

The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are -890.3 KJ mol^-1, – 393.5 KJ mol^-1 and – 285.8 KJ mol^-1 respectively. Enthalpy of formation of CHJg) will be
(i) – 74.8 KJ mol^-1  (ii) – 52.27 KJ mol^-1
(iii) + 74.8 KJ mol^-1 (iv) + 52.26 KJ mol^-1

 A reaction, A + B—>C + D + q is found to have a positive entropy change. The reaction will be
(i) possible at high temperature (ii) possible only at low temperature
(iii) not possible at any temperature (iv) possible at any temperature

 In a process, 701 ] of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process?

The reaction of cyanamide,NH₂CN(s) with dioxygen was carried out in a bomb calorimeter and ∆U was found to be -742,7 KJ^-1  mol^-1 at 298 K. Calculate the enthalpy change for the reaction at 298 K.NH₂CN (S) + 3/202(g) —–>N₂(g) + CO₂(g) + H₂0(Z)

 Calculate the number of kj of heat necessary to raise the temperature of 60 g of aluminium from 35°C to 55°C. Molar heat capacity of Al is 24 J mol^-1 K^-1.

 Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at – 10.0°C. A, H = 6.03 KJ mot1 at 0°C. Cp [H20(l)J = 75.3 J mol^-1 K^-1; Cp [H20(s)J = 36.8 J mol^-1 K^-1.

 Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at – 10.0°C. A, H = 6.03 KJ mot1 at 0°C. Cp [H20(l)J = 75.3 J mol^-1 K^-1; Cp [H20(s)J = 36.8 J mol^-1 K^-1.

 Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at – 10.0°C. A, H = 6.03 KJ mot1 at 0°C. Cp [H20(l)J = 75.3 J mol^-1 K^-1; Cp [H20(s)J = 36.8 J mol^-1 K^-1.

Given : N₂(g) + 3H₂(g) ————> 2NH₃(g); ∆_{r H^–} = -92.4 kj mot^-1 What is the standard enthalpy of formation of NH₃ gas?