Which of the following statements is correct for the spontaneous adsorption of a gas?
1. | ∆ S is negative and therefore, ∆ H should be highly positive |
2. | ∆ S is negative and therefore, ∆ H should be highly negative |
3. | ∆ S is positive and therefore, ∆ H should be negative |
4. | -∆ S is positive and therefore, ∆ H should also be highly positive |
Given the following reaction:
\(4H(g)\)→ \(2 H_{2}\)\((g)\)
The enthalpy change for the reaction is -869.6 kJ. The dissociation energy of the H-H bond is:
1. -869.6 kJ
2. +434.8kJ
3. +217.4kJ
4. -434.8 kJ
From the following bond energies:
H—H bond energy: 431.37 kJ mol-1
C=C bond energy: 606.10 kJ mol-1
C—C bond energy: 336.49 kJ mol-1
C—H bond energy: 410.50 kJ mol-1
Enthalpy for the reaction,
will be:
1. | 1523.6 kJ mol-1 | 2. | -243.6 kJ mol-1 |
3. | -120.0 kJ mol-1 | 4. | 553.0 kJ mol-1 |
The values of ΔH and ΔS for the given reaction are 170 kJ and 170 JK-1, respectively.
C(graphite) + CO2(g)→2CO(g)
This reaction will be spontaneous at:
1. 710 K
2. 910 K
3. 1110 K
4. 510 K
1. 1.968 V
2. 2.0968 V
3. 1.0968 V
4. 0.0968 V
The bond energy of H—H and Cl-Cl is 430 kJ mol-1 and 240 kJ mol-1 respectively
and ΔHf for HCl is -90 kJ mol-1. The bond enthalpy of HCl is:
1. 290
2. 380
3. 425
4. 245
Two moles of an ideal gas is heated at a constant pressure of one atmosphere from 27C to 127C. If then q and for the process are respectively:
1. 6362.8 J, 4700 J
2. 3037.2 J, 4700 J
3. 7062.8 J, 5400 J
4. 3181.4 J, 2350 J
Calculate (in kJ/mol) for from the and the values provided at 27C
1. -2258.1 kJ/mol
2. -1129.05 kJ/mol
3. -964.35 kJ/mol
4. None of the above
The correct statement for a reversible process in a state of equilibrium is:
1. G = – 2.30RT log K
2. G = 2.30RT log K
3. Go = – 2.30RT log K
4. Go = 2.30RT log K