The heat of combustion of carbon to CO2 is –393.5 KJ/mol. The heat released upon the formation of 35.2 g of CO2 from carbon and oxygen gas is:
| 1. | –315 KJ | 2. | +315 KJ |
| 3. | –630 KJ | 4. | +630 KJ |
| 1. | 38 | 2. | 317 |
| 3. | 477 | 4. | 32 |
The enthalpy of fusion of water is 1.435 kcal/mol. The molar entropy change for the melting of ice at 0 oC is:
| 1. | 10.52 cal/(mol K) | 2. | 21.04 cal/(mol K) |
| 3. | 5.260 cal/(mol K) | 4. | 0.526 cal/(mol K) |
Give 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 following reaction will be:

Consider the following reactions:
| (i) | H+(aq) + OH-(aq) → H2O(l) ΔH = -x1 kJmol-1 |
| (ii) | H2(g) + 1/2O2(g) → H2O(l) ΔH = -x2 kJmol-1 |
| (iii) | CO2(g) + H2(g) → CO (g) + H2O(l) ΔH = -x3 kJmol-1 |
| (iv) | C2H2(g) + 5/2O2(g) → 2CO2 + H2O(l) ΔH = -x4 kJmol-1 |
Enthalpy of formation of H2O(l) is :
1.
2.
3.
4.
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
The enthalpies of combustion of H₂, cyclohexene (C₆H₁₀), and cyclohexane (C₆H₁₂) are −241 kJ mol⁻¹, −3800 kJ mol⁻¹, and −3920 kJ mol⁻¹, respectively. Calculate the heat of hydrogenation of cyclohexene for the reaction:
C₆H₁₀ + H₂ → C₆H₁₂
1. -121 kJ per mol
2. +121 kJ per mol
3. +242 kJ per mol
4. -242 kJ per mol
If the bond energies of H–H, Br – Br, and H –Br are 433, 192, and 364 kJ mol–1 respectively, the ∆Hº for the reaction H2(g) + Br2(g) → 2HBr(g) will be:
| 1. | +103 kJ | 2. | +261kJ |
| 3. | –103 kJ | 4. | –261 kJ |
The formation of a solution from two components can be considered as:
| (i) | Pure solvent → separated solvent molecules, ∆H1 |
| (ii) | Pure solute → separated solute molecules, ∆H2 |
| (iii) | Separated solvent and solute molecules → solution, ∆H3 |
The solution so formed will be ideal if:
1. ∆HSoln = ∆H1 + ∆H2 + ∆H3
2. ∆HSoln = ∆H1 + ∆H2 – ∆H3
3. ∆HSoln = ∆H1 – ∆H2 – ∆H3
4. ∆HSoln = ∆H3 – ∆H1 – ∆H2