3H2(g)+N2(g)=2NH3(g)
Suppose the equilibrium constant Kp = 0.003337 for the reaction above. If the equilibrium mixture contains partial pressures H2 = 0.400 atm and N2 = 0.350 atm. What is the equilibrium partial pressure of NH3 in atm?
Kp = pNH3^2/pN2*pH2^3
You have only one unknown. Solve for that.
To find the equilibrium partial pressure of NH3 (denoted as P(NH3)), we can use the equilibrium constant expression and the given partial pressures of H2 and N2.
The balanced equation for the reaction is: 3H2(g) + N2(g) = 2NH3(g)
The equilibrium constant expression in terms of partial pressures (Kp) can be written as:
Kp = (P(NH3))^2 / (P(H2))^3 * P(N2)
Given that Kp = 0.003337, P(H2) = 0.400 atm, and P(N2) = 0.350 atm, we can rearrange the expression to solve for P(NH3).
0.003337 = (P(NH3))^2 / (0.400)^3 * 0.350
Multiplying both sides of the equation by (0.400)^3 * 0.350, we get:
0.003337 * (0.400)^3 * 0.350 = (P(NH3))^2
Simplifying the right side of the equation, we have:
0.003337 * 0.064 * 0.350 = (P(NH3))^2
0.00008968 = (P(NH3))^2
Taking the square root of both sides, we find:
P(NH3) ≈ ±0.009469 atm
Since partial pressures cannot be negative, we take the positive value of P(NH3):
P(NH3) ≈ 0.009469 atm
Therefore, the equilibrium partial pressure of NH3 is approximately 0.009469 atm.