What is the change in Gibbs' free energy for $"C"(s) + "O"_2(g) -> "CO"_2(g) + "393.51 kJ/mol"$ ?

Question

The $∆S^@$ for the reaction,

$"C"(s) + "O"_2(g) -> "CO"_2(g) + "393.51 kJ/mol"$ ,

at $"308.62 K"$ is $"0.004 kJ/mol"cdot"K"$ .

Find the value of $∆G^@$ at $"308.62 K"$ .

What would be the answer in $"kJ/mol"$ ? Thanks!

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Answer 1 · 2018-03-18T04:47:30

Follow the steps below

Look at the reaction and determine the $DeltaH^@$
Since the reaction is exothermic: $DeltaH^@= -393.51 ("kJ")/("mol")$
Given $DeltaS^@ = 0.004 ("kJ")/("mol"*"K")$
Given $T= "308.62 K"$
Use the formula: $DeltaG^@=DeltaH^@-TDeltaS^@$
Plug in givens: $DeltaG^@=-393.51 ("kJ")/("mol")-(308.62 cancel("K"))*(0.004 ("kJ")/("mol"*cancel("K")))$

$= -394.74 ("kJ")/("mol")$

Answer 2 · 2018-03-18T05:08:25

$-394.74$ $" kJ"/"mol"$

To solve this question, use the standard-state free energy of reaction equation:

$Delta"G"_0=DeltaH_0 - TDeltaS_0$

Given that you were given all the values for enthalpy, temperature, and entropy:

$Delta"G"_0=(-393.51 " kJ"/"mol") - (308.62" K")(0.004 "kJ"/("mol" *"K"))$

$Delta"G"_0= -394.74$ $" kJ"/"mol"$

Since the value Delta $"G"_0$ is negative, this reaction is thermodynamically favorable.