A) \[427\text{ }KJ\]
B) \[245\text{ }KJ\]
C) \[285\text{ }KJ\]
D) \[766\,KJ\]
Correct Answer: B
Solution :
\[\frac{1}{2}{{H}_{2}}+\frac{1}{2}C{{l}_{2}}\xrightarrow{{}}HCl\] \[\Delta H=91\,KJ\,mo{{l}^{-1}}\] \[\Delta {{H}_{recation}}=\left[ \frac{1}{2}BE({{H}_{2}})+\frac{1}{2}BE(C{{l}_{2}}) \right]-BE\,(HCl)\] \[91=\left[ \frac{1}{2}\times 430+\frac{1}{2}\times 242 \right]-BE(HCl)\] \[BE(HCl)=(215+121)-91=245\,KJ\,mo{{l}^{-1}}\]You need to login to perform this action.
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