ThermochemistryHard
Question
The enthalpy of atomization of PH3(g) is +954 kJ/ mol and that of P2H4 is +1.488 MJ/mol. The bond energy of the P–P bond is
Options
A.318 kJ/mol
B.372 kJ/mol
C.216 kJ/mol
D.534 kJ/mol
Solution
$PH_{3}(g) \rightarrow P(g) + 3H(g);\Delta H = 954\text{ kJ}$
$\therefore 3 \times B.E._{P - H} = 954 \Rightarrow B.E._{P - H} = 318\text{ kJ/mol} $$${P_{2}H_{4}(g) \rightarrow 2P(g) + 4H(g);\Delta H = 1488\text{ kJ} }{\therefore B.E._{P - P} + 4 \times B.E._{P - H} = 1488 \Rightarrow B.E._{P - P} = 216\text{ kJ/mol}}$$
Create a free account to view solution
View Solution FreeMore Thermochemistry Questions
When 1.0 g of oxalic acid is burnt in a bomb calorimeter whose heat capacity is 8.75 kJ/K, the temperature increases by ...2MnO4 – + 16H+ + 10Cl– → 2Mn2+ + 5Cl2(g) + 8H2OThe above reaction is endothermic and hence, the actual temperature of th...A solution of 500 ml of 2 M-KOH is added to 500 ml of 2 M-HCl and the mixture is well shaken. The rise in temperature T1...Which of the following forms oxide on heating :-...Among the following, for which reaction the heat of reaction represents bond energy of HCl?...