Question

1. A student ran the following reaction in the laboratory at 582 K:
CO(g) + Cl₂(g)      COCl₂(g)
When she introduced 0.380 moles of CO(g) and 0.403 moles of Cl₂(g) into a 1.00 liter container, she found the equilibrium concentration of COCl₂(g) to be 0.342 M.
Calculate the equilibrium constant, K_c, she obtained for this reaction.
2. The equilibrium constant, K_c, for the following reaction is 1.29×10^-3 at 540 K.
COCl₂(g) CO(g) + Cl₂(g)
When a sufficiently large sample of COCl₂(g) is introduced into an evacuated vessel at 540 K, the equilibrium concentration of Cl₂(g) is found to be 0.349 M.
Calculate the concentration of COCl₂ in the equilibrium mixture.

3. Write the equilibrium constant expression, K_c, for the following reaction:

Please enter the compounds in the order given in the reaction.

If either the numerator or denominator is 1, please enter 1

NH₃(g) + H₂S(g) NH₄HS(s)

4. Write the equilibrium constant expression, K_c, for the following reaction:

If either the numerator or denominator is 1, please enter 1

Cr³⁺(aq) + 3OH^-(aq) Cr(OH)₃(s)

Answer 1

1)

K_c = 147.5

Explanation

CO(g) + Cl2(g) <-------> COCl2(g)

K_c = [COCl2]/([CO][Cl2])

Initial concentration

[CO] = 0.380

[Cl2] = 0.403

[COCl2] = 0

Change in concentration

[CO] = - x

[Cl2] = - x

[COCl2] = +x

Equilibrium concentration

[CO] = 0.380 - x

[Cl2] = 0.403 - x

[COCl2] = x

at equilibrium ,

[COCl2] = 0.342M

x = 0.342

[CO] = 0.380 - 0.342 = 0.038M

[Cl2] = 0.403 - 0.342 = 0.061M

K_c = 0.342M/( 0.038M × 0.061M)

K_c = 147.5

2)

94.4M

Explanation

COCl2(g) <-------> CO(g) + Cl2(g)

K_c = [CO][Cl2]/[COCl2] = 1.29 ×10^-3

at equilibrium

[CO] = [Cl2] = 0.349M

[COCl2] = [CO][Cl2]/K_c

[COCl2] = 0.349M × 0.349M/( 1.29 ×10^-3)

[COCl2] = 94.4M

3)

K_c = 1/([NH3][H2S])

4)

K_c = 1/([Cr³⁺][OH^-]³)