The decomposition of ammonia is: 2 NH₃(g)⇌ N₂(g)+ 3 H₂(g).If K_p 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?

Question

Quizplus · Accepted Answer

The equilibrium constant expression for the decomposition of ammonia ($K_p$) is given by $K_p = \frac{[N_2][H_2]^3}{[NH_3]^2}$. Given $K_p = 1.5 \times 10^3$, $P_{N_2} = 0.10$ atm, and $P_{H_2} = 0.15$ atm, we can solve for $P_{NH_3}$. Rearranging the equation to solve for $P_{NH_3}$ and substituting the given values, we find that $P_{NH_3}$ equals approximately $4.7 \times 10^{-4}$ atm.

The Decomposition of Ammonia Is: 2 NH3(g)⇌ N2(g)+ 3 H2(g)

The Decomposition of Ammonia Is: 2 NH₃(g)⇌ N₂(g)+ 3 H₂(g)