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Thermodynamics question

2017 · Shift 0 · Q18

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JEE Main
Chemistry
Thermodynamics
2017 · Shift 0 · Q18

Thermodynamics question

2017 · Shift 0 · Q18

JEE MainChemistryThermodynamicsMCQ+4 / −1

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 :

A
+144.0 kJ mol–1
B
– 74.8 kJ mol–1
C
-144.0 kJ mol–1
D
+ 74.8 kJ mol–1

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