JEE MAIN - Chemistry (2017 (Offline) - No. 21)
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 :
$${\Delta _r}{H^o}$$ = - 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
Explanation
C(graphite) + O2(g) $$ \to $$ CO2 (g);
$${\Delta _r}{H^o}$$ = - 393.5 kJ mol-1 .............(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 .........(2)
$$C{O_2}(g)$$ + $$2{{\rm H}_2}{\rm O}(l) \to$$ $$C{H_4}(g)$$ + $$2{O_2}(g)$$
$${\Delta _r}{H^o}$$ = + 890.3 kJ mol-1 .........(3)
C(graphite) + 2H2(g) $$ \to $$ CH4 (g); $$\Delta $$H = ? ........ (4)
You can see,
[Eq. (1) + Eq. (3)] + [2 × Eq. (2)] = Eq. (4)
$$ \therefore $$ [ $$\Delta $$rH1 + $$\Delta $$rH3 ] + [2 $$ \times $$ $$\Delta $$rH2] = $$\Delta $$H
$$ \Rightarrow $$ $$\Delta $$H = [(-393.5) + (890.3)] + [2(-285.8)] = -74.8 kJ / mol
$${\Delta _r}{H^o}$$ = - 393.5 kJ mol-1 .............(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 .........(2)
$$C{O_2}(g)$$ + $$2{{\rm H}_2}{\rm O}(l) \to$$ $$C{H_4}(g)$$ + $$2{O_2}(g)$$
$${\Delta _r}{H^o}$$ = + 890.3 kJ mol-1 .........(3)
C(graphite) + 2H2(g) $$ \to $$ CH4 (g); $$\Delta $$H = ? ........ (4)
You can see,
[Eq. (1) + Eq. (3)] + [2 × Eq. (2)] = Eq. (4)
$$ \therefore $$ [ $$\Delta $$rH1 + $$\Delta $$rH3 ] + [2 $$ \times $$ $$\Delta $$rH2] = $$\Delta $$H
$$ \Rightarrow $$ $$\Delta $$H = [(-393.5) + (890.3)] + [2(-285.8)] = -74.8 kJ / mol
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