InterviewSolution
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InterviewSolution
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Given `C_(("graphite"))+O_(2)(g)toCO_(2)(g),` `Delta_(r)H^(0)=-393.5kJ" "mol^(-1)` `H_(2)(g)=+(1)/(2)O_(2)(g)toH_(2)O(1),` `Delta_(r)H^(0)=-285.8" kJ "mol^(-1)` `CO_(2)(g)+2H_(2)O(1)toCH_(4)(g)+2O_(2)(g)`, `Delta_(r)H^(0)=+890.3kJ" "mol^(-1)` Based on the above thermochemical equations, the value of `Delta_(r)H^(0)` at at 298 K for the reaction `C_(("graphite"))+2H_(2)(g)toCH_(4)(g)` will be:A. `+74.8kJ" "mol^(-1)`B. `+144.0kJ" "mol^(-1)`C. `-74.8kJ" "mol^(-1)`D. `-144.8kJ:" " mol^(-1)` |
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Answer» Correct Answer - C From the given equations: `C_(("graphite"))+O_(2)(g)toCO_(2)(g),Delta_(r)H^(@)=-393.5Kj" "Mol^(-1)` `2H_(2)(g)+O_(2)(g)to2H_(2)O(l),Delta_(r)H^(@)=-285.8xx2kJ" "mol^(-1)` `CO_(2)(g)+2H_(2)O(l)toCH_(4)(g)+2O_(2)(g),Delta_(r)H^(@)=+890.3kJ" "mol^(-1)` Adding above three equations we get `C_(("Graphite"))+2H_(2)(g)toCH_(4)(g)` `Delta_(r)H^(@)=-393.5-285.8xx2+890.3` `=74.8kJ" "mol^(-1)`. |
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