Question

5. Calculate \(\Delta G^\circ\) for the following reaction at 298 K. Cu(s) + S(g) \to CuS(s) A) 291.8 kJ B) -184.6 kJ C) 184.6 kJ D) -291.8 kJ \begin{tabular}{|l|c|c|c|} \hline & \(\Delta G^\circ_f\) & \(\Delta H^\circ_f\) & \(S^\circ\) \\ & (kJ/mol) & (kJ/mol) & (J/mol\cdot K) \\ \hline Cu(s) & 0 & 0 & 33.2 \\ S(g) & 238.2 & 278.8 & 167.8 \\ CuS(s) & -53.6 & -53.1 & 66.6 \\ \hline \end{tabular}

          5. Calculate \(\Delta G^\circ\) for the following reaction at 298 K.
Cu(s) + S(g) \to CuS(s)
A) 291.8 kJ
B) -184.6 kJ
C) 184.6 kJ
D) -291.8 kJ
\begin{tabular}{|l|c|c|c|}
\hline
 & \(\Delta G^\circ_f\) & \(\Delta H^\circ_f\) & \(S^\circ\) \\
 & (kJ/mol) & (kJ/mol) & (J/mol\cdot K) \\
\hline
Cu(s) & 0 & 0 & 33.2 \\
S(g) & 238.2 & 278.8 & 167.8 \\
CuS(s) & -53.6 & -53.1 & 66.6 \\
\hline
\end{tabular}

5. Calculate Δ G^∘ for the following reaction at 298 K.
Cu(s) + S(g) →CuS(s)
A) 291.8 kJ
B) -184.6 kJ
C) 184.6 kJ
D) -291.8 kJ

Δ G^ Δ H^ S^∘

(kJ/mol) (kJ/mol) (J/mol·K)

Cu(s) 0 0 33.2

S(g) 238.2 278.8 167.8

CuS(s) -53.6 -53.1 66.6

Added by Tanya A.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Ajay Singhal

08/09/2021

Step 1

Given: Cu(s): 0 kJ/mol S(g): 0 kJ/mol CuS(s): -53.6 kJ/mol \Delta G\deg = \sum n \Delta G\deg_f(products) - \sum m \Delta G\deg_f(reactants) \Delta G\deg = (-53.6) - (0 + 0) \Delta G\deg = -53.6 kJ/mol Show more…

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Calculate Delta Gdeg f for the following reaction at 298K. Cu(s)+S(g)->CuS(s) A) 291.8kJ B) -184.6kJ C) 184.6kJ D) -291.8kJ 5. Calculate Gf for the following reaction at 298 K Cu(s)+ S(g) > CuS(s) Gf H&f so (kJ/mol) (kJ/mol) (J/mol-K) 0 0 33.2 238.2 278.8 167.8 -53.6 -53.1 66.6 A) 291.8 kJ B) -184.6 kJ C) 184.6 kJ D) -291.8 kJ Cu(s) s(g) CuS(s)