The equilibrium constant, Kp, for the following reaction is 0.160 at 298K. 2NOBr(g) 2NO(g) + Br2(g) If an equilibrium mixture of the three gases in a 15.9 L container at 298K contains NOBr at a pressure of 0.363 atm and NO at a pressure of 0.421 atm, the equilibrium partial pressure of Br2 is atm. Consider the following reaction:
2NH3(g) N2(g) + 3H2(g) If 3.53 * 10-4 moles of NH3, 0.297 moles of N2, and 0.320 moles of H2 are at equilibrium in a 16.6 L container at 945 K, the value of the equilibrium constant, Kp, is .

1. The partial pressure of Br₂ is 0.118951 atm 2) The value of Kp is 2.843 × 10².

Explanation:

1. The reaction at equilibrium will be:

2NOBr (g) ⇒ 2NO (g) + Br₂ (g)

The partial pressure of PNOBr is 0.363 atm, the partial pressure of NO is 0.421 atm, there is a need to find the partial pressure of Br₂, the value of Kp is 0.160. For the given reaction,

Kp = [PNO]² × [PBr₂] / [PNOBr]²

0.160 = [0.421]² × [PBr₂] / [0.363]²

0.160 = [0.177241 × [PBr₂] / 0.131769

[PBr₂] = 0.131769 × 0.160 / 0.177241

[PBr₂] = 0.118951 atm

The partial pressure of Br₂ is 0.118951 atm.

2) The reaction will be,

2NH₃ ⇔ N2 (g) + 3H₂ (g)

Kp is the equilibrium constant, and its value will be,

Kp = [PN₂] × [PH₂]³ / [PNH₃]² (i)

Considering the reactants in terms of molarity will be,

Kp = [N2] × [H₂]³ / [NH₃]²

Here, [N2] = 0.297 / 16.6 = 0.017892 M

[H₂] = 0.320/16.6 = 0.019277 M

[NH₃] = 3.53 × 10⁻⁴ / 16.6 = 2.12651 × 10⁻⁵ M

Now putting the values in equation (i) we get,

Kp = [0.017892] × [7.16399 × 10⁻⁶] / [4.52204 × 10⁻¹⁰]

Kp = 0.128167 × 10⁻⁶ / 4.52204 × 10⁻¹⁰

Kp = 2.8343 × 10²

Hence, the equilibrium constant is 2.8343 × 10²

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