Given, ${C_{(graphite)}} + {O_2} \to C{O_2}(g)$; ${\Delta _r}{H^o}$ = - 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 :

Question

Given, ${C_{(graphite)}} + {O_2} \to C{O_2}(g)$;

${\Delta _r}{H^o}$ = - 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 :

+144.0 kJ mol^–1

– 74.8 kJ mol^–1

-144.0 kJ mol^–1

+ 74.8 kJ mol^–1

Solution

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