The following reaction has Kp = 109 at 25°C.
2 NO(g) + Br2(g) 2 NOBr(g)
If the equilibrium partial pressure of Br2 is 0.0159 atm and the equilibrium partial pressure of NOBr is 0.0768 atm, calculate the partial pressure of NO at equilibrium.
To calculate the partial pressure of NO at equilibrium, we can use the equation and the given equilibrium constant (Kp).
The balanced equation for the reaction is: 2 NO(g) + Br2(g) ⇌ 2 NOBr(g)
According to the given information, the equilibrium partial pressure of Br2 is 0.0159 atm and the equilibrium partial pressure of NOBr is 0.0768 atm.
Using the equilibrium partial pressures, we can set up an expression for the equilibrium constant (Kp):
Kp = (P_NOBr)^2 / (P_NO)^2 * (P_Br2)
Where:
Kp is the equilibrium constant
P_NOBr is the partial pressure of NOBr
P_NO is the partial pressure of NO
P_Br2 is the partial pressure of Br2
We can substitute the given values:
109 = (0.0768)^2 / (P_NO)^2 * (0.0159)
Now, let's solve for the partial pressure of NO (P_NO):
109 = (0.0768)^2 / (P_NO)^2 * (0.0159)
Rearranging the equation:
(P_NO)^2 = ((0.0768)^2 * 0.0159) / 109
Taking the square root of both sides:
P_NO = √(((0.0768)^2 * 0.0159) / 109)
Calculating this value:
P_NO ≈ 0.0323 atm
Therefore, the partial pressure of NO at equilibrium is approximately 0.0323 atm.
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K = (NOBr)^2/(NO)^2(Br2)
You have only one unknown. Solve for it.