A) 214
B) 426
C) 318
D) 1272
Correct Answer: A
Solution :
\[P{{H}_{3}}\left( g \right)\xrightarrow{{}}p\left( g \right)+3H\left( g \right)\] |
\[\Delta H=953\,kJ\,mo{{l}^{-1}}\] |
BE of \[\left( P-H \right)bond=\frac{953}{3}\] |
\[{{P}_{2}}{{H}_{4}}\left( g \right)\xrightarrow{{}}2P\left( g \right)+4H\left( g \right)\] |
This involve breaking of one \[(P-P)\] bond and four \[(P-H)\] bonds. |
\[\therefore \,\,\Delta H=\frac{953}{3}\times 4+BE\left( P-P \right)\] |
\[1485=\frac{953}{3}\times 4+BE\left( p-p \right)\] |
\[BE\left( P-P \right)=214.3kJ\] |
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