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
The enthalpies of formation of FeO(s) and Fe2O3(s) are −65.0 and −197.0 kcal/mol, respectively. A mixture of the two oxi...The enthalpy change in a reaction does not depend upon...One gram sample of NH4NO3 is decomposed in a bomb calorimer, the temperature of the calorimeter increase by 6.12 K. The ...The reaction of zinc metal with hydrochloric acid was used to produce 1.5 moles of hydrogen gas at 298 K and 1 atm press...Among the following, for which reaction the heat of reaction represents bond energy of HCl?...