Question

A 1.00 kg sample of Sb2S3 (s) and a 10.0 g sample of H2 (g) are allowed to react in a 25.0 L container at 713 K. At equilibrium, 72.6 g H2S (g) is present? What is the value of K at 713 Kelvin for the following reaction?

Answer 1

Answer: The value of $K_c$ is coming out to be 0.412

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$     .....(1)

• For $Sb_2S_3$

Given mass of $Sb_2S_3$ = 1.00 kg = 1000 g   (Conversion factor: 1 kg = 1000 g)

Molar mass of $Sb_2S_3$ = 339.7 g/mol

Putting values in equation 1, we get:

$\text{Moles of }Sb_2S_3=\frac{1000g}{339.7g/mol}=2.944mol$

• For hydrogen gas:

Given mass of hydrogen gas = 10.0 g

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1, we get:

$\text{Moles of hydrogen gas}=\frac{10.0g}{2g/mol}=5mol$

• For hydrogen sulfide:

Given mass of hydrogen sulfide = 72.6 g

Molar mass of hydrogen sulfide = 34 g/mol

Putting values in equation 1, we get:

$\text{Moles of hydrogen sulfide}=\frac{72.6g}{34g/mol}=2.135mol$

The chemical equation for the reaction of antimony sulfide and hydrogen gas follows:

                  $Sb_2S_3(s)+3H_2(g)\rightarrow 2Sb(s)+3H_2S(g)$

Initial:            2.944      5

At eqllm:      2.944-x     5-3x         2x        3x

We are given:

Equilibrium moles of hydrogen sulfide = 2.135 moles

Calculating for 'x', we get:

$\Rightarrow 3x=2.135\\\\\Rightarrow x=\frac{2.135}{3}=0.712$

Equilibrium moles of hydrogen gas = (5 - 3x) = (5 - 3(0.712)) = 2.868 moles

Volume of the container = 25.0 L

Molarity of a solution is calculated by using the formula:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume}}$

The expression of $K_c$ for above equation, we get:

$K_c=\frac{[H_2S]^3}{[H_2]^3}$

The concentration of solids and liquids are not taken in the expression of equilibrium constant.

$K_c=\frac{(\frac{2.135}{25})^3}{(\frac{2.868}{25})^3}\\\\K_c=0.412$

Hence, the value of $K_c$ is coming out to be 0.412