Given, C (graphite) + O₂ → C O₂ (g); Δ r H^(o) = - 393.5 kJ mol⁻¹ H₂(g) + 1/2O₂(g) → H₂ O(l) Δ r H^(o) = - 285.8 kJ mol⁻¹ C O₂ (g) + 2 H₂ O(l) → C H₄ (g) + 2O₂(g) Δ r H^(o) = + 890.3 kJ mol⁻¹ Based on the above thermochemical equations, the value of Δ r H^(o) at 298 K for the reaction C (graphite) + 2 H₂(g) → C H₄ (g) will be:

Chemical Thermodynamics: JEE Main 2017 Chemistry question

JEE Main 2017 · Session, Shift 1 · Q2349

JEE Main 2017Session, Shift 1ChemistryHardChemical Thermodynamics84× since 2002

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 :

JEE Main 2017 Chemical Thermodynamics Question with Solution (Session Shift 1, Q2349) | JEE Arc