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.
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To calculate the partial pressure of NO at equilibrium, we can use the expression of the equilibrium constant (Kp) and the given equilibrium partial pressures of Br2 and NOBr.
The equation for the reaction is:
2 NO(g) + Br2(g) -> 2 NOBr(g)
The equilibrium constant expression for this reaction is:
Kp = (P_NOBr)^2 / (P_NO)^2 * (P_Br2)
Given:
Kp = 109
P_Br2 = 0.0159 atm
P_NOBr = 0.0768 atm
Let's substitute these values into the equilibrium constant expression to solve for P_NO:
109 = (0.0768)^2 / (P_NO)^2 * (0.0159)
Cross multiplying the equation:
109 * (P_NO)^2 * (0.0159) = (0.0768)^2
Dividing both sides by (0.0159):
109 * (P_NO)^2 = (0.0768)^2 / (0.0159)
Taking the square root of both sides:
P_NO = √[(0.0768)^2 / (0.0159 * 109)]
Calculating this expression:
P_NO ≈ √(0.00591 / 1.731) ≈ √0.0034159914 ≈ 0.0585
Therefore, the partial pressure of NO at equilibrium is approximately 0.0585 atm.
To calculate the partial pressure of NO at equilibrium, we can use the equation for the equilibrium constant Kp and apply it to the given values. The equation for Kp in this case is as follows:
Kp = (P(NOBr)^2) / (P(NO)^2 * P(Br2))
Given:
Kp = 109
P(Br2) = 0.0159 atm
P(NOBr) = 0.0768 atm
Let's substitute these values into the equation:
109 = (0.0768^2) / (P(NO)^2 * 0.0159)
Now, let's solve for P(NO):
P(NO)^2 = (0.0768^2) / (109 * 0.0159)
P(NO) = √((0.0768^2) / (109 * 0.0159))
P(NO) ≈ √(0.00590032 / 0.001731)
P(NO) ≈ √3.407
P(NO) ≈ 1.844 atm
Therefore, the partial pressure of NO at equilibrium is approximately 1.844 atm.