The equilibrium constant for the following reaction is 1.6 × 10⁵ at 1024K H₂(g) + Br₂(g) ⇌ 2HBr(g) Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024K.

The equilibrium pressures are: p H 2 = 0.025 bar, p Br 2 = 0.025 bar, and p HBr = 9.95 bar. — Since the equilibrium constant Kₚ is very large (1.6 × 10⁵), the reaction heavily favors the formation of HBr. When only HBr is introduced, a very small amount will dissociate into H₂ and Br₂ to establish equilibrium. By setting up an ICE table in terms of pressure and substituting into the Kₚ expression, we can solve for the change in pressure and find the final equilibrium pressures.

The equilibrium constant for the following reaction is 1.6 × 10⁵ at…

Question 27 of 69intermediate🔧 ApplyNumerical3 marks

The equilibrium constant for the following reaction is $1.6 \times 10^5$ at 1024K

H_2(g) + Br_2(g) \rightleftharpoons 2HBr(g)

Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024K.

Correct Answer

The equilibrium pressures are: $p_{H_2} = 0.025$ bar, $p_{Br_2} = 0.025$ bar, and $p_{HBr} = 9.95$ bar.

Exercise: EXERCISES | Q: 6.27 | (Page 44)

For More Understanding

Explanation

Since the equilibrium constant $K_p$ is very large ( $1.6 \times 10^5$ ), the reaction heavily favors the formation of HBr. When only HBr is introduced, a very small amount will dissociate into $H_2$ and $Br_2$ to establish equilibrium. By setting up an ICE table in terms of pressure and substituting into the $K_p$ expression, we can solve for the change in pressure and find the final equilibrium pressures.

Solution Steps

Step 1: Let the reaction proceed in the backward direction to reach equilibrium. Let $2x$ be the decrease in pressure of HBr.
$2HBr(g) \rightleftharpoons H_2(g) + Br_2(g)$
Initial pressure: $10.0$ bar, $0$ , $0$ Equilibrium pressure: $(10.0 - 2x)$ bar, $x$ bar, $x$ bar
Step 2: Write the expression for $K_p$ for the forward reaction $H_2(g) + Br_2(g) \rightleftharpoons 2HBr(g)$ .
$K_p = \frac{(p_{HBr})^2}{(p_{H_2})(p_{Br_2})}$
Step 3: Substitute the equilibrium pressures into the $K_p$ expression.
$1.6 \times 10^5 = \frac{(10.0 - 2x)^2}{(x)(x)}$
Since $K_p$ is very large, $x$ will be very small, so $(10.0 - 2x) \approx 10.0$ .
$1.6 \times 10^5 = \frac{(10.0)^2}{x^2}$ $x^2 = \frac{100}{1.6 \times 10^5} = \frac{1}{1600}$
$x = \sqrt{\frac{1}{1600}} = 0.025$ bar
Step 4: Calculate the equilibrium pressures of all gases. $p_{H_2} = x = 0.025$ bar $p_{Br_2} = x = 0.025$ bar $p_{HBr} = 10.0 - 2x = 10.0 - 2(0.025) = 10.0 - 0.05 = 9.95$ bar

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