Question

Consider following reactions:

I. $\left[Co\right(N{H}_3{)}_6{]}^{3+}\left(aq\right)+6H_3{O}^{+}\left(aq\right)\rightleftharpoons \left[Co\right(H_{2}O{)}_6{]}^{3+}\left(aq\right)+6N{H}_4^{+}\left(aq\right)$

The products are favoured in the above reaction:

II. $\left[Co\right(N{H}_3{)}_6{]}^{2+}\left(aq\right)+6H_3{O}^{+}\left(aq\right)\to \left[Co\right(H_{2}O{)}_6{]}^{2+}\left(aq\right)+6N{H}_4^{+}\left(aq\right)$

The reaction (II) takes place with $H_{2}O$ in a matter of seconds.

Which is true in terms of $\Delta G^o$?

• A. $\Delta G_I^o$ is more negative than that of $\Delta G_{II}^o$
• B. $\Delta G_I^o$ is less negative than that of $\Delta G_{II}^o$
• C. $\Delta G_I^o=\Delta G_{II}^o<0$
• D. $\Delta G_I^o=\Delta G_{II}^o>0$

Answer 1

Answer: (B)

$\Delta G_I^o$ is less negative than that of $\Delta G_{II}^o$

$\Delta G_I^o$ is less negative than that of II.

$\left[Co\right(N{H}_3{)}_6{]}^{3+}$ is thermodynamically unstable but kinetically stable relative to $\left[Co\right(H_{2}O{)}_6{]}^{3+}$. The former has a lower value of formation constant so its $\Delta G_I^o$ is also less negative than that of latter.

Therefore, option (B) is correct.