Question

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? 7) A) 4.7 × 10-4 atm B) 2.2 × 10-7 atm C) 2.1 × 103 atm D) 4.4 × 106 atm

Answer 1

Answer:

The partial pressure of NH3 at the equilibrium is 4.7 * 10^-4 atm (option A )

Explanation:

Step 1: Data given

Kp = 1.5 * 10³ at 400 °C

Partial pressure of N2 = 0.10 atm

Partial pressure of H2 = 0.15 atm

Step 2: The balanced equation

2 NH3(g) → N2(g) + 3 H2(g)

Step 3: Calculate partial pressure of NH3

Kp = (pN2)*(pH2)³ / (pNH3)²

⇒ with Kp = 1.5*10³

⇒with pN2 = the partial pressure of N2 = 0.10 atm

⇒with pH2 = the partial pressure of H2 = 0.15 atm

⇒ with pNH3 = the partial pressure of NH3 = TO BE DETERMINED

1500 = (0.10*0.15³) / (pNH3)²

(pNH3)² = 0.000000225

pNH3 = 0.00047 atm = 4.7 * 10^10^-4 atm

The partial pressure of NH3 at the equilibrium is 4.7 * 10^-4 atm (option A )

Answer 2

Answer:

A) = 4.7 × 10⁻⁴atm

Explanation:

Given that,

Kp = 1.5*10³ at 400°C

partial pressure pN2 = 0.10 atm

partial pressure pH2 = 0.15 atm

To determine:

Partial pressure pNH3 at equilibrium

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

$K_p = \frac{[pH_2] ^3[pN_2]}{[pNH_3]^2} \\pNH_3 = \sqrt{\frac{(pH_2)^3(pN_2)}{pNH_3} } \\pNH_3 = \sqrt{\frac{(0.15)^3(0.10)}{1.5 \times 10^3} } \\=4.74 \times 10^-^4atm$

= 4.7 × 10⁻⁴atm