Question

question 5
use the bond enthalpies given below to estimate the enthalpy of reaction, in kj/mol, for the following reaction:
ph₃(g) + 3 hcl(g) → pcl₃(g) + 3 h₂(g)
question 6

Explanation:

Step1: Calculate the energy required to break bonds

The reactant \(PH_3\) has \(3\) \(H - P\) bonds and \(3\) \(HCl\) has \(3\) \(H - Cl\) bonds.
The energy to break bonds: \(3\times326+3\times431.9\) \(kJ/mol\)

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Step2: Calculate the energy released when forming bonds

The product \(PCl_3\) has \(3\) \(P - Cl\) bonds and \(3\) \(H_2\) has \(3\) \(H - H\) bonds.
The energy released when forming bonds: \(3\times328+3\times436.4\) \(kJ/mol\)

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Step3: Calculate the enthalpy of reaction

Enthalpy of reaction \(\Delta H=\) (Energy to break bonds) - (Energy released when forming bonds)

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Answer:

\(-19.5\space kJ/mol\)