12H2(g)→H(g) ΔHf=50 kcal/mol PH3(g)→P(g)+3H(g)ΔHdissociation=24 – InterviewSolution

1.	12H2(g)→H(g) ΔHf=50 kcal/mol PH3(g)→P(g)+3H(g)ΔHdissociation=243 kcal/mol P - P bond dissociation energy is x×10 kcal/mol. The value of x in nearest possible integer is
Answer» $\frac{1}{2} H_{2} (g) \to H (g) Δ H_{f} = 50 kcal/mol$ $P H_{3} (g) \to P (g) + 3 H (g) Δ H_{dissociation} = 243 kcal/mol$ P - P bond dissociation energy is $x \times 10 kcal/mol$ . The value of $x$ in nearest possible integer is

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