The decomposition of ammonia is: 2 NH3 (g) ? N2 (g) + 3 H2 (g). If Kp is 1.5 * 103 at 400°C, what is the partial pressure of ammonia at equilibrium when N2 is 0.10 atm and H2 is 0.15 atm? The decomposition of ammonia is: 2 NH3 (g) ? N2 (g) + 3 H2 (g). If Kp is 1.5 * 103 at 400°C, what is the partial pressure of ammonia at equilibrium when N2 is 0.10 atm and H2 is 0.15 atm? 4.7 * 10-4 atm 2.2 * 10-7 atm 2.1 * 103 atm 4.4 * 106 atm

Question

Keyon Coleman

Chemistry

16 May, 04:51

The decomposition of ammonia is: 2 NH3 (g) ? N2 (g) + 3 H2 (g). If Kp is 1.5 * 103 at 400°C, what is the partial pressure of ammonia at equilibrium when N2 is 0.10 atm and H2 is 0.15 atm? The decomposition of ammonia is: 2 NH3 (g) ? N2 (g) + 3 H2 (g). If Kp is 1.5 * 103 at 400°C, what is the partial pressure of ammonia at equilibrium when N2 is 0.10 atm and H2 is 0.15 atm? 4.7 * 10-4 atm 2.2 * 10-7 atm 2.1 * 103 atm 4.4 * 106 atm

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Answers (1)

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Shelby Sparks · Answer 1

Given:

Kp = 1.5*10³

partial pressure pN2 = 0.10 atm

partial pressure pH2 = 0.15 atm

To determine:

Partial pressure pNH3 at equilibrium

Explanation:

The decomposition reaction is:-

2NH3 (g) ↔N2 (g) + 3H2 (g)

Kp = [pH2]³[pN2]/[pNH3]²

pNH3 = √ [ (pH2) ³ (pN2) / Kp]

pNH3 = √ (0.15) ³ (0.10) / 1.5*10³ = 4.74*10⁻⁴ atm

Ans (a)