Question: $\Delta G_{F}^{\circ}$ for the formation of HI (g) from its gaseous elements is –9.67 KJ/mol at 50...

Question

$\Delta G_{F}^{\circ}$ for the formation of HI (g) from its gaseous elements is –9.67 KJ/mol at 500K. When the partial pressure of HI is 10.0 atm and of I₂ is 0.001 atm. What must be the partial pressure of hydrogen at this temperature to reduce the magnitude of DG for the reaction to zero –

Answer 1

Solution

$\frac{1}{2}$ H₂ + $\frac{1}{2}$ I₂ ¾® HI, $\Delta G_{f}^{\circ}$ = – 9.67 KJ/mol

If DG = 0, then K_eq = Q. (Q = reaction quotient)

Now from $\Delta G_{f}^{\circ}$ = –RT ln K_eq

Ž 9.67 = $\frac{8.314 \times 500}{1000}$ lnK_eq

Ž K_eq = 10

Ž 10 = $\frac{P_{HI}}{\left( P_{H_{2}} \right)^{1/2}\left( P_{I_{2}} \right)^{1/2}} \Rightarrow P_{H_{2}} = \frac{100}{100 \times 10^{–3}}$

= 10³ atm

Question: \(\Delta G_{F}^{\circ}\) for the formation of HI (g) from its gaseous elements is –9.67 KJ/mol at 50...

Solution