InterviewSolution
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InterviewSolution
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The enthalpy change for the following processes are listed below `{:(Cl_(2)(g)=2 Cl(g),242.3 kJ mol^(-1)),(I_(2)(g)=2I(g),151.0 kJ mol^(-1)),(ICl(g)=I(g)+Cl(g),211.3 kJ mol^(-1)),(I_(2)(s)=I_(2)(g),62.76 kJ mol^(-1)):}` Given that the standard states for iodine and chlorine are `I_(2)(g)` and `Cl_(2)(g)`, the standard enthalpy of formation for ICl(g) isA. `-14.6 kJ mol^(-1)`B. `-16.8 kJ mol^(-1)`C. `+16.8 kJ mol^(-1)`D. `+244.8 kJ mol^(-1)` |
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Answer» Correct Answer - C `1//2I_(2)(s)+1//2Cl_(2)(g)rarrICl(g), DeltaH_(f)=?` `{:(Cl_(2)(g)rarr2Cl(g),,DeltaH=242.3 kJmol^(-1)),(I_(2)(s)rarrI_(2)(g),,DeltaH=62.76 kJ mol^(-1)),(I_(2)(g)rarr2I(g),,DeltaH=151.0 kJ mol^(-1)):}` `2Cl(g)+2I(g)rarr2ICl(g), DeltaH=-422.6 kJ mol^(-1)` `I_(2)(s)+Cl_(2)(g)rarr2ICl(g),` `thereforeH=-422.6 +151+62.76 +242.3=33.46 kJ` So heat of formation of `ICl=(33.46)/(2)=16.73 kJ mol^(-1)`. |
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