Chemical EquilibriumHard

Question

The Deacon reaction is the oxidation of HCl by O2 in the following reaction

HCl(g) + ¼ O2(g) $\rightleftharpoons$½ Cl2(g) + ½ H2O(g) at a pressure of 730 mm and with an initial mixture containing 8% of HCl and 92% of O2, the degree of decomposition of the HCl is 0.08. What is the equilibrium partial pressure of oxygen?

Options

A.671.6 mm
B.659.92 mm
C.537.28 mm
D.670.43 mm

Solution

HCl(g) + ¼ O2(g) $\rightleftharpoons$ ½ Cl2(g) + ½ H2O(g)

Initial partial pressure $730 \times \frac{8}{100}$ $730 \times \frac{92}{100}$

= 58.4 mm = 671.6 mm

Equilibrium partial pressure 58.4 – 58.4 × 0.08

$671.6 - \frac{58.4 \times 0.08}{4} = 670.432\text{ mm}$

Create a free account to view solution

View Solution Free
Topic: Chemical Equilibrium·Practice all Chemical Equilibrium questions

More Chemical Equilibrium Questions

In a vessel of 1.0 L capacity, O2(g) at 0.25 atm pressure and HCl(g) at 1.0 atm pressure are allowed to react in the pre...ΔG° for the reaction X + Y$\rightleftharpoons$C is – 4.606 kcal at 1000 K. The equilibrium constant for the reverse mode...The equilibrium constant for the reaction w + x ⇋ y + z is 9. If one mole of each of w and x are mixed and there i...The progress of the reaction A $\rightleftharpoons$nB with time is represented by the graph given below. The value of n ...A quantity of 34 g sample of BaO2 is heated to 1000 K in a closed and rigid evacuated vessel of 8.21 L capacity. What pe...