Chemical EquilibriumHard
Question
ΔfGo for the formation of HI(g) from its gaseous elements is – 2.303 kcal/mol at 500 K. When the partial pressure of HI is 10 atm and of I2(g) is 0.001 atm, then what must be the partial pressure of hydrogen be at this temperature to reduce the magnitude of ΔG for the reaction to zero?
Options
A.1000 atm
B.10000 atm
C.100 atm
D.31.63 atm
Solution
$\Delta G^{o} = - 2.303RT.\ln K_{eq}$
$\Rightarrow - 2.303 \times 10^{3} = - 2.303 \times 2 \times 500 \times \ln K_{eq} $$${\therefore K_{eq} = 10 }{\text{Now, }K_{eq} = \frac{P_{HI}}{P_{H_{2}}^{1/2} \times P_{I_{2}}^{1/2}} \Rightarrow \frac{10}{P_{H_{2}}^{1/2} \times (0.001)^{1/2}}}$$
$\therefore P_{H_{2}} = 1000\text{ atm}$
Create a free account to view solution
View Solution FreeMore Chemical Equilibrium Questions
The value of $\frac{K_{C}}{K_{P}}$ the below for reaction is CO(g) + ½ O2(g) $\rightleftharpoons$ CO2(g)...For the equilibrium at 298 K; N2O4(g) ⇋ 2NO2(g); G-N2O4 = 100 kJ mol-1 and G-NO2 = 50 kJ mol-1. If 5 mol of N2O4 a...Two gases A and B, one being the dimer of the other, are at equilibrium. Decreasing of pressure at constant temperature ...Consider the following equilibrium in a closed container. N2O4(g) $\rightleftharpoons$2NO2(g)At a fixed temperature, the...Some quantity of water is contained in a container as shown in figure. As neon is added to this system at constant press...