Consider this reaction:

$I_2(g)+C{l}_2(g)2$ICl$(g),K=6\mathrm{.}91{10}^3$

$at460K$

${}_{r}G=-33\mathrm{.}8\frac{kJ}{\text{m}\text{o}\text{l}}$

Calculate ${}_{r}G$ for the reaction at $460K$ under

each of the following conditions:

Correct

Part B

at equilibrium

Express your answer using three significant figures and include the appropriate units.

Previous Answers

Correct

Part C

$P_{I_2}=\frac{0\mathrm{.}325}{b}ar$

$P_{C{l}_2}=0\mathrm{.}499$ bar

$P_{\text{I}\text{C}\text{l}}=2\mathrm{.}55$ bar

Express your answer using one decimal place and include the appropriate units.