Question

The equilibrium constant, K. for the following reaction is 2.20x104 at 723K 2NH3(8) N2(g) + 3H2(8) If an equilibrium mixture of the three gases in a 13.1 L container at 723K contains NH, at a pressure of 0.693 atm and Ny at a pressure of 1.02 atm, the equilibrium partial pressure of His atm. Submit Answer Retry Entire Group 9 more group attempts remaining

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

, 2NH₃ (g) <--------------> N₂(g) + 3H₂(g)

Given, P_NH3  = 0.693 atm , P_N2 = 1.02 atm , P_H2 = we need to calculate

Step 3: Calculation of K_P

2NH₃ (g) <--------------> N₂(g) + 3H₂(g)

we know, The K_p expression can be written as

K_p = ( P³_H2 × P_N2 ) / P²_NH3  

on substituting the values of partial pressure and K_p = 2.20  × 10⁴

=> 2.20  × 10⁴ = ( P³_H2 × 1.02) / (0.693 )²

=> P³_H2 = ( 2.20  × 10⁴ × (0.693 )²  ) / 1.02 = 10358.31176

=>  P_H2 = ∛10358.33176 = 21.8 atm [ note:  if K_p = 2.20  × 10⁴ then P_H2 = 21.8 atm ]

[ note: if K_p = 2.20  × 10^-4 then P_H2 = 0.047 atm ]