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Calculate the standard free energy chang...

Calculate the standard free energy change for the formation of methane at `300K`:
`C("graphite") +2H_(2) (g) rarr CH_(4)(g)`
The following data are given:
`Delta_(f)H^(Theta) (kJ mol^(-1)): CH_(4)(g) =- 74.81`
`Delta_(f)S^(Theta)(JK^(-1) mol^(-1)): C("graphite") = 5.70, H_(2)(g) = 130.7 CH_(4)(g) = 186.3`

A

`Delta_(r)G^(@)` is -50.8 kJ and the reactionis driven by enthalpy only

B

`Delta_(r)G^(@)` is`-50.8 kJ` and the reaction is driven by entropy only

C

`Delta_(r)G^(@)` is `+5.0 kJ` and the reaction is driven by enthalpy and entropy

D

`Delta_(r)G^(@)` is `-50.8 kJ` and the reaction is driven by enthalpy and entropy

Text Solution

Verified by Experts

The correct Answer is:
A
`Delta_(r)H = Delta_(f)_(CH_(4))^(@) = -74.9 kJ//mol` and `Delta_(r)S^(@) = S_(CH_(4))^(@) - 2S_(H_(2))^(@) - S_(C)^(@) = 186.3 -2 xx 130.7 - 5.6 = -80.7 J//mol`
`Delta_(r)G^(@) = Delta_(r)H^(@) - Tdelta_(r)S^(@) =-74.9 - 298 xx (-80.7) xx 10^(-3) = -50.85` kJ/mol
`rArr` Reaction is spontaneous due to highly negative enthalpy change
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