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With the help of thermochemical equation...

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 CO_(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)

Text Solution

Verified by Experts

The required equation is
(i) `C("graphite") + 2H_(2)(g) to CH_(4)(g) Delta_(r)H^(@) = ?`
Other equations given are :
(ii)`C("graphite") + O_(2)(g) to CO_(2)(g) Delta_(r)H^(@) = -393.5 kJ mol^(-1)`
(iii) `H_(2)(g) + 1/2O_(2)(g) to H_(2)O(l) Delta_(r)H^(@) = -285.8 kJ mol^(-1)`
(iv) `CO_(2)(g) + 2H_(2)O(l) to CH_(4)(g) + 2O_(2)(g) Delta_(r)H^(@) = + 890.3 kJ mol^(-1)`
Multiply eq.(iii) by 2 and add to eq.(ii)
(ii) `C("graphite") + O_(2)(g) to CO_(2)(g) Delta_(r)H^(@) = -393.5 kJ mol^(-1)`
(iii)` xx 2 2h_(2)(g) + O_(2)(g) to 2H_(2)O(l) Delta_(r)H^(@) = -571.6 kJ mol^(-1)`
Adding
(v) `C("graphite") + 2H_(2)(g) + 2O_(2)(g) to CO_(2)(g) + 2H_(2)O(l) Delta_(r)H^(@) = -965.1 kJ mol^(-1)`
Add eqn (iv) to (v)
(iv) `CO_(2)(g) + 2H_(2)O(l) to CH_(4)(g) + 2O_(2)(g) Delta_(r)H^(@) = + 890.3 kJ mol^(1)`
Adding `C("graphite") + 2H_(2)(g) to CH_(4)(g) Delta_(r)H^(@) = -74.8 kJ mol^(-1)`
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