Consider the reaction below at 298 K : C (graphite) + 2H 2 (g) → CH 4 (g) ∆ f Hº (kJ/mol) = 74.9 S 0 m (J/K/mol) + 5.6 + 130.7 + 186.3 Which statement below is correct ?

Consider the reaction below at 298 K : C (graphite) + 2H 2 (g) → CH 4 (g) ∆ f Hº (kJ/mol)…

Chemistry · JEE Advanced · NTA Exams — Chemical Thermodynamics

Consider the reaction below at 298 K :
C (graphite) + 2H₂ (g) → CH₄ (g) ∆_fHº (kJ/mol) = 74.9
S⁰_m(J/K/mol)
+ 5.6 + 130.7 + 186.3
Which statement below is correct ?

∆_rG⁰ is – 50.8 kJ and the reaction is driven by enthalpy only.
∆_rG⁰ is – 50.8 kJ and the reaction is driven by entropy
only.
∆_rG⁰ is + 50.8 kJ and the reaction is driven by both
enthalpy and entropy.
∆_rG⁰ is – 50.8 kJ and the reaction is driven by both
enthalpy and entropy.

Answer

(A) ∆ r G 0 is – 50.8 kJ and the reaction is driven by enthalpy only.

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Consider the reaction below at 298 K : C (graphite) + 2H 2 (g) → CH 4 (g) ∆ f Hº (kJ/mol)…

Answer

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