At 2200°C, K = 0.050 for the following reaction.
N2(g) + O2(g) 2 NO(g)
What is the partial pressure of NO in equilibrium with N2 and O2 that were placed in a flask at initial pressures of 0.75 atm and 0.30 atm, respectively?
is K of 0.050 Kp or Kc?
it didn't specify in the question. I think its just a constant.
To determine the partial pressure of NO in equilibrium, we can use the expression for the equilibrium constant (K) and the initial partial pressures of N2 and O2.
The equilibrium constant expression for the reaction is:
K = (P_NO^2) / (P_N2 * P_O2)
Where P_NO, P_N2, and P_O2 are the partial pressures of NO, N2, and O2, respectively.
In this case, we are given that K = 0.050 and the initial pressures of N2 and O2 are 0.75 atm and 0.30 atm, respectively.
To find the partial pressure of NO, we can rearrange the equilibrium constant expression and solve for P_NO:
P_NO^2 = K * (P_N2 * P_O2)
P_NO^2 = 0.050 * (0.75 atm * 0.30 atm)
P_NO^2 = 0.01125 atm^2
Taking the square root of both sides to solve for P_NO:
P_NO = √(0.01125 atm^2)
P_NO ≈ 0.106 atm
Therefore, the partial pressure of NO in equilibrium is approximately 0.106 atm.