Question

The following reaction has Kp = 109 at 25°C. 2 NO(g) + Br2(g) equilibrium reaction arrow 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.

Answer 1

Answer: The equilibrium partial pressure of NO is 0.0034 atm

Explanation:

For the given chemical equation:

$NO(g)+Br_2(g)\rightleftharpoons 2NOBr(g)$

The expression of $K_p$ for above equation follows:

$K_p=\frac{(p_{NOBr})^2}{p_{NO}\times p_{Br_2}}$

We are given:

Equilibrium partial pressure of $Br_2$ = 0.0159 atm

Equilibrium partial pressure of NOBr = 0.0768 atm

$K_p=109$

Putting values in above equation, we get:

$109=\frac{(0.0768)^2}{p_{NO}\times 0.0159}\\\\p_{NO}=0.0034atm$

Hence, the equilibrium partial pressure of NO is 0.0034 atm