Consider the following reaction at 298 K: C(graphite) + 2H2(g)—> CH4(g) Delta ∆ H = -74..6 kJ?

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Calculate the following quantities.

Delta ∆ S (sys) —– j/K

Delta ∆ S (sur) —– j/K

Delta ∆ S (univ) —– j/K

C Graphite = 5.7 S° J/(mol·K)

2H2(g) = 130.7 S° J/(mol·K)

CH4(g) = 186.3 S° J/(mol·K)

I do not know how to find the answer, my chemistry teacher is horrible at teaching. Please help frustrated from college kid.

1 Answer

  • Calculate the ΔS for the system:

    Sum of entropy of products – sum of entropy of reactants

    186.3 – [2(130.7) + 5.7 ]

    186.3 – [ 261.4 + 5.7 ]

    186.3 – 267.1 = -80.8 Joules/K

    Entropy change of surroundings: Since the reaction evolves heat, the entropy change

    of the surrounding is a positive value.

    Delta H = -74.6 kilojoules x 1000 Joules/ kilojoule = -74,600 Joules

    ΔS = ΔH / T = +74,600 Joules/ 298.15 K = +250.2 Joules/K

    Entropy change of universe: -80.8 Joules + 250.2 Joules = +169.4 Joules/K

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