InterviewSolution
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InterviewSolution
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The enthalpy changes for the following process are listed below : `Cl_(2)(g)=2Cl(g)," "242.3" kJ"mol^(-1)` `I_(2)(g)=2I(g)," "151.0" kJ"mol^(-1)` `ICl(g)=2I(g)+Cl(g)," "211.3" kJ"mol^(-1)` `I_(2)(s)=I_(2)(g)," "62.76" kJ"mol^(-1)` Given that standard states for iodine and chlorine are `I_(2)(s)` and `Cl_(2)(g)`, the standerd enthalpy of formation for ICl(g) is :A. `+16.8 kJ mol^(-1)`B. `+244.8 kJ mol^(-1)`C. `-14.6 kJ mol^(-1)`D. `-16.8 kJ mol^(-1)` |
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Answer» Correct Answer - A `1/2 I_(2)+1/2 Cl_(2) rarr ICl, DeltaH=H_(R)-H_(P)` |
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