Question

please answer all parts and show work.

20) The equilibrium constant for e reaction Ni2+ (aa) H3 (ag) a Ni(NH3)62+ (ng) is th K-5.6x10s at 298 K. a) What is AG0 at this temperature?

Answer 1

Part A)

Using the following equation we get the dG

$\Delta G^{0}=-RTln(K)$

$\Delta G^{0}=-(8.314\frac{Pam3}{molK})(298K)ln(5.6x10^{8})$

$\Delta G^{0}= -49,906.8 J/mol$

Part B

As dG<<0, the process is spontaneous, therefore the reaction at standart conditions of P and T is product favored.

Part C

We need to get K

$K=\frac{[Ni(NH3)6]}{[Ni][NH3]^{6}}$

$K=\frac{(0.010)}{(0.0010)(0.0050)^{6}}=6.4X10^{14}$

Now using the equation [1] we find dG

$\Delta G^{0}=-(8.314)(298K)ln(6.4x10^{14})$

$\Delta G^{0}= -84,466.6 J/mol$

Part D

I cant' see the sketch. Please upload a photo with that part of the problem.