Question

Given :

calculate the equilibrium constant for $\text{H}{\text{g}_{2}}^{\text{2+}} \rightarrow \text{Hg + H}\text{g}^{\text{2+}}\text{ .}$

• A. 3.13 × 10-3
• B. 3.13 × 10-4
• C. 6.26 × 10-3
• D. 6.26 × 10-4

Answer 1

Answer: (C)

6.26 × 10-3

Answer 2

$\text{+ 2}\text{e}^{-}\ \overset{\quad\quad}{\rightarrow}\text{2Hg , 0.789 VoltHg }\overset{\quad\quad}{\rightarrow}\text{ H}\text{g}^{\text{2+}}\text{ + 2}\text{e}^{-}\text{ , -0.854 VoltH}{\text{g}_{2}}^{\text{2+}}\ \overset{\quad\quad}{\rightarrow}\text{Hg + H}\text{g}^{\text{2+}}\text{ , -0.065 Volt}\Delta\text{ G = - 2}\text{×}\text{ }(\text{- 0.065})\text{ }\text{×}\text{ 96500 = -8.314 }\text{×}\text{ 298 ln }\text{K}_{\text{eq.}}\text{ ;}\text{K}_{\text{eq.}}\text{= 6.3 }\text{×}\text{ 1}\text{0}^{\text{-3}}$