The decomposition of ammonia is: 2 NH₃(g)⇌ N₂(g)+ 3 H₂(g).If the pressure of ammonia is 1.0 × 10^-3 atm,and the pressures of N₂ and H₂ are each 0.20 atm,what is the value for K_p' at 400°C for the reverse reaction?

Question

Quizplus · Accepted Answer

The equilibrium constant for the reverse reaction, $K_p'$, can be calculated using the given pressures of ammonia ($NH_3$), nitrogen ($N_2$), and hydrogen ($H_2$). The expression for $K_p'$ for the reverse reaction is the inverse of the $K_p$ for the forward reaction, which is given by $\frac{P_{NH_3}^2}{P_{N_2} \times P_{H_2}^3}$. Substituting the given pressures, we get $K_p' = \frac{(1.0 \times 10^{-3})^2}{(0.20) \times (0.20)^3} = 6.2 \times 10^{-4}$.

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)