Question

04 Question (1 point) A reaction vessel contains an equilibrium mixture of SO2, O2, and SO3. The reaction proceeds such that: 2802(g) +0,(8) — 2803(e) 6th attempt The partial pressures at equilibrium are: Pso, = 0.001915 atm Po, = 0.001111 atm Pso, = 0.0166 atm Calculate Kp for the reaction. x 104 - 3 OF 16 QUESTIONS COMPLETED < 04/16 > + VIEW SOLUTION

Answer 1

Answer:

Step 1: Explanation

Kp is the equilibrium constant calculated from the partial pressures of a reaction equation. It is used to express the relationship between product pressures and reactant pressures

Step 2: Write the balanced the chemical equation

2SO₂ (g) + O₂(g) <--------------> 2SO₃(g)

Given, P_SO2 =0.001915 atm , P_O2 = 0.001111 atm , P_SO3 = 0.0166 atm

Step 3: Calculation of K_P

2SO₂ (g) + O₂(g) <--------------> 2SO₃(g)

we know, The K_p expression can be written as

K_p = P²_SO3 / ( P²_SO2  × P_O2 )

on substituting the values of partial pressure

=> K_p = ( 0.0166 )² / ( (0.001915 )² × 0.001111 ) = 6.76  × 10⁴ [ exact value = 6.763392017 × 10⁴ ]