Respuesta :
The question is incomplete, here is the complete question:
Sulfur dioxide and oxygen react to form sulfur trioxide during one of the key steps in sulfuric acid synthesis. An industrial chemist studying this reaction fills 125 L a tank with 60 moles of sulfur dioxide gas and 12 moles of oxygen gas at 30°C and when the mixture has come to equilibrium measures the amount of sulfur trioxide gas to be 12 moles.
Calculate the concentration equilibrium constant for the reaction of sulfur dioxide and oxygen at the final temperature of the mixture. Round your answer to 2 significant digits.
Answer: The concentration equilibrium constant for the reaction is 1.302
Explanation:
We are given:
Initial moles of sulfur dioxide = 60 moles
Initial moles of oxygen gas = 12 moles
Equilibrium moles of sulfur trioxide = 12 moles
Volume of tank = 125 L
Molarity is calculated by using the equation, we get:
[tex]\text{Initial molarity of sulfur dioxide}=\frac{60}{125}=0.48M[/tex]
[tex]\text{Initial molarity of oxygen gas}=\frac{12}{125}=0.096M[/tex]
[tex]\text{Equilibrium molarity of sulfur trioxide}=\frac{12}{125}=0.096M[/tex]
The chemical equation for the reaction of sulfur dioxide and oxygen gas follows:
[tex]2SO_2+O_2\rightleftharpoons 2SO_3[/tex]
Initial: 0.48 0.096
At eqllm: 0.48-2x 0.096-x 2x
Evaluating the value of 'x':
[tex]\Rightarrow 2x=0.096\\\\x=0.048[/tex]
So, equilibrium concentration of sulfur dioxide = (0.48 - 2x) = [0.48 - 2(0.048)] = 0.384 M
Equilibrium concentration of oxygen = (0.096 - x) = [0.096 - 0.048] = 0.048 M
The expression of [tex]K_c[/tex] for above equation follows:
[tex]K_c=\frac{[SO_3]^2}{[O_2]\times [SO_2]^2}[/tex]
Putting values in above equation, we get:
[tex]K_c=\frac{(0.096)^2}{0.048\times (0.384)^2}\\\\K_c=1.302[/tex]
Hence, the concentration equilibrium constant for the reaction is 1.302
