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With the help of thermochemical equations given below, determine `Delta_(r )H^(Θ)` at `298 K` for the following reaction: `C("graphite")+2H_(2)(g) rarr CH_(4)(g),Delta_(r )H^(Θ) = ?` `C("graphite")+O_(2)(g) rarr CH_(2)(g), Delta_(r )H^(Θ) = -393.5 kJ mol^(-1)` ...(1) `H_(2)(g) +1//2O_(2)(g) rarr H_(2)O(l)`, `Delta_(r )H^(Θ) = -285.8 kJ mol^(-1)` ...(2) `CO_2(2)(g)+2H_(2)O(l) rarr CH_(4)(g)+2O_(2)(g)`, `Delta_(r )H^(Θ) = +890.3 kJ mol^(-1)` ...(3) |
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Answer» Here we want one mole of `C_(("graphite"))` as reactant, so we write down eauation (i), we want two mole of `H_(2)(g)` as reactant, so we multiply equation (ii) by `2`, we want one mole of `CH_(4)(g)` as product, so we write down equation (iii) as such, `C_(("graphite"))+O_(2)(g) rarr CO_(2)(g), Delta_(r )H^(Θ)=-393.5 kJ mol^(-1)` ...(i) `2H_(2)(g)+O_(2) rarr 2H_(2)O(l), Delta_(r)H^(Θ)=2(-285.8 kJ mol^(-1))` ...(ii) `CO_(2)(g)+2H_(2)O(l) rarr CH_(4)(g)+2O_(2)(g), Delta_(r)H^(Θ)=+890.3 kJ mol^(-1)` ...(iii) Adding we obtain `C_(("graphite"))+2H_(2)(g) rarr CH_(4)(g), Delta_(r)H^(Θ)=-74.8 kJ mol^(-1)` |
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