What is K_c for the following equilibrium when the equilibrium concentration of each substance is: [SO₂]= 0.60 M, [O₂] = 0.82 M and [SO₃] = 1.90 M ?
\(2SO_2(g) + O_2(g) ⇋ 2SO_3(g)\)

Question

For the reaction \( \mathrm{N_2} + 3\mathrm{H_2} \rightleftharpoons 2\mathrm{NH_3} \), the equilibrium constant \(K_c\) is 0.5 at a certain temperature. If initially, 1 mole of \(\mathrm{N_2}\) and 3 moles of \(\mathrm{H_2}\) are placed in a 1 L container, what is the equilibrium concentration of \(\mathrm{NH_3}\)?

Answer 1

Given:

Equilibrium reaction:
2SO₂(g) + O₂(g) ⇌ 2SO₃(g)

Equilibrium concentrations:
[SO₂] = 0.60 M
[O₂] = 0.82 M
[SO₃] = 1.90 M

Step 1: Write the equilibrium constant expression

K_c = [SO₃]² / ( [SO₂]² [O₂] )

Step 2: Substitute the given values

K_c = (1.90)² / ( (0.60)² × 0.82 )

K_c = 3.61 / (0.36 × 0.82)

K_c = 3.61 / 0.2952

K_c = 12.23

Final Answer:

The equilibrium constant for the reaction is:
K_c ≈ 12.2

What is K_c for the following equilibrium when the equilibrium concentration of each substance is: [SO₂]= 0.60 M, [O₂] = 0.82 M and [SO₃] = 1.90 M ?
\(2SO_2(g) + O_2(g) ⇋ 2SO_3(g)\)

Solution and Explanation

Top Questions on Law Of Chemical Equilibrium And Equilibrium Constant

What is Kc for the following equilibrium when the equilibrium concentration of each substance is: [SO2]= 0.60 M, [O2] = 0.82 M and [SO3] = 1.90 M ?\(2SO_2(g) + O_2(g) ⇋ 2SO_3(g)\)

Solution and Explanation

Top Questions on Law Of Chemical Equilibrium And Equilibrium Constant

What is K_c for the following equilibrium when the equilibrium concentration of each substance is: [SO₂]= 0.60 M, [O₂] = 0.82 M and [SO₃] = 1.90 M ?
\(2SO_2(g) + O_2(g) ⇋ 2SO_3(g)\)