ThermochemistryHard
Question
From the following data at 25o C, which of the following statement(s) is/are correct?
½H2(g) + ½O2(g)→ OH(g): ΔHo = 42 kJ
H2(g) + ½O2(g) → H2O(g): ΔHo = −242 kJ
H2(g) → 2H(g): ΔHo = 436 kJ
O2(g) →2O(g): ΔHo = 495 kJ
Options
A.ΔrHo for the reaction H2O(g) → 2H(g) + O(g) is 925.5 kJ.
B.ΔrHo for the reaction OH(g) → H(g) + O(g) is 502 kJ.
C.Enthalpy of formation of H(g) is −218 kJ/mol.
D.Enthalpy of formation of OH(g) is 42 kJ/mol.
Solution
$(a)\Delta_{r}H = \left( 436 + \frac{1}{2} \times 495 \right) - ( - 242) = 925.5\text{ kJ}$
$(b)\Delta_{r}H = \left( \frac{1}{2} \times 436 + \frac{1}{2} \times 495 \right) - ( - 42) = 423.5\text{ kJ} $$${(c)\Delta_{r}H_{H(g)} = \frac{1}{2} \times 436 = 218\text{ kJ/mol} }{(d)\Delta_{f}H_{OH(g)} = 42\text{ kJ/mol}}$$
Create a free account to view solution
View Solution FreeMore Thermochemistry Questions
The enthalpy of atomization of graphite is 698.6 kJ/mol and the mean bond enthalpy of C−C bond in diamond is 348.4 kJ/mo...When 1.0 g of oxalic acid (H2C2O4) is burned in a bomb calorimeter whose heat capacity is 8.75 kJ/K, the temperature inc...Formation of ozone from oxygen is an endothermic process. In the upper atmosphere, ultraviolet is the source of energy t...Calculate ΔfH for ZnSO4(s) from the following dataZnS(s) → Zn(s) + S (rhombic), ΔH1 = 44 kcal/mol2ZnS(s) + 3O2(g) → 2ZnO...For the given two processes(i) ½P4(s) + 3Cl2(g) → 2PCl3(l): ΔH = −635 kJ (ii) PCl3(l) + Cl2(g) → PCl5(s): ΔH = −137 kJth...