Thermodynamics · Question 64

The densities of graphite and diamond at 298 K are 2.25 and 3.31 g cm-3, respectively. If the standard free energy difference ( G ) is equal to 1895 J mol-1, the pressure at which graphite will be transformed into diamond at 298 K is

1. 11.14 108 Pa
2. 11.14 107 Pa
3. 11.14 106 Pa
4. 11.14 105 Pa

What is the entropy change (in J K-1 mol-1) when one mole of ice is converted into water at 0 C? (The enthalpy change for the conversion of ice to liquid water is 6.0 kJ mol-1 at 0 C).

1. 20.13
2. 2.013
3. 2.198
4. 21.98

For which one of the following equations is H react equal to H f for the product ?

1. Xe(g) + 2F2(g) XeF4(g)
2. 2CO(g) + O2(g) 2CO2(g)
3. N2(g) + O3(g) N2O3(g)
4. CH4(g) + 2Cl2(g) CH2Cl2(l) + 2HCl(g)

The molar heat capacity of water at constant pressure, C, is 75 J K-1 mol-1. When 1.0 kJ of heat is supplied to 100 g of water which is free to expand, the increase in temperature of water is

1. 1.2 K
2. 2.4 K
3. 4.8 K
4. 6.6 K

Unit of entropy is

1. J K-1 mol-1
2. J mol-1
3. J-1 K-1 mol-1
4. J K mol-1

In a closed insulated container a liquid is stirred with a paddle to increase the temperature. Which of the following is true?

1. E = W 0, q = 0
2. E = W = q 0
3. E = 0, W = q 0
4. W = 0, E = q 0

Heat of combustion Ho for C(s), H2(g) and CH4(g) are -94, -68 and -213 kcal/mol, then Ho for C(s) + 2H2(g) CH4(g) is

1. -17 kcal
2. -111 kcal
3. -170 kcal
4. -85 kcal

2 moles of an ideal gas at 27 temperature is expanded reversibly from 2 lit. to 20 lit. Find entropy change. (R = 2 cal/mol K)

1. 92.1
2. 0
3. 4
4. 9.2

Change in enthalpy for reaction, 2H2O2(l) 2H2O(l) + O2(g) if heat of formation of H2O2(l) and H2O(l) are -188 and -286 kJ/mol respectively, is

1. -196 kJ/mol
2. +196 kJ/mol
3. +948 kJ/mol
4. -948 kJ/mol

When 1 mol of gas is heated at constant volume temperature is raised from 298 to 308 K. Heat supplied to the gas is 500 J. Then which statement is correct ?

1. q = w = 500 J, E = 0
2. q = E = 500 J, w = 0
3. q = -w = 500 J, E = 0
4. E = 0, q = w = -500 J