Question

Consider the reaction below and its K_p value at 350K:

H_2(g)   +   Br_2(g)   <--->   2HBr_(g)       K_p = 3.5 x 10⁴

If the partial pressures of H₂ = 0.024 atm and HBr = 5.07 atm, what is the equilibrium partial pressure of Br₂?

	a.	12 atm
	b.	4.7 atm
	c.	0.26 atm
	d.	0.031 atm

Answer 1

Solution:

The Equilibrium Reaction given here is:

H2(g) + Br2(g) -----> 2HBr(g).

Partial Pressure of H2(g) (pH2) = 0.024atm.

Partial Pressure of HBr(g) (pHBr)= 5.07atm.

The Equilibrium Constant of the Reaction (Kp) = 3.4×10^4.

Consider a Reaction; aX + bY -----> cZ

If pX ,pY and pZ is the Partial Pressures of the Gases and a,b and c is the Number of Moles of the Gases,then;

The Equilibrium Constant of this Reaction is related to the Partial Pressure of the Gases as:

Kp = (pZ)^c ÷ [(pX)^a × (pY)^b].

Then the Equilibrium Constant of the given Reaction is Calculated as follows:

Kp = (pHBr)^2 ÷ [pH2 × pBr2].

Then; 3.4×10^4 = 5.07^2 ÷ (0.024 atm × pBr2).

pBr2 = 25.7 ÷ (0.024× 3.4×10^4).

Therefore, the Equilibrium Partial Pressure of Br2 (pBr2) = 0.031 atm.

So Option d. is the Correct Answer.

Note: Please do like the Answer.