JEE MAIN - Chemistry (2017 (Offline))
- 21Given, $${C_{(graphite)}} + {O_2} \to C{O_2}(g)$$;
$${\Delta _r}{H^o}$$ = - 393.5 kJ mol-1
$${{\rm H}_2}(g)$$ + $${1 \over 2}{O_2}(g)$$$$\to {{\rm H}_2}{\rm O}(l)$$
$${\Delta _r}{H^o}$$ = - 285.8 kJ mol-1
$$C{O_2}(g)$$ + $$2{{\rm H}_2}{\rm O}(l) \to$$ $$C{H_4}(g)$$ + $$2{O_2}(g)$$
$${\Delta _r}{H^o}$$ = + 890.3 kJ mol-1
Based on the above thermochemical equations, the value of $${\Delta _r}{H^o}$$ at 298 K for the reaction
$${C_{(graphite)}}$$ + $$2{{\rm H}_2}(g) \to$$ $$C{H_4}(g)$$ will be :Одговорити(B)– 74.8 kJ mol–1

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