A sample of O2 with an initial temperature of 50.0 oC and a volume of 105 L is cooled to -25 oC. The new pressure is 105.4 kPa a
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<u>Answer:</u> The concentration of carbon dioxide, hydrogen gas, carbon monoxide and water when equilibrium is re-established are 0.362 M, 0.212 M, 0.138 M and 0.138 M respectively.
<u>Explanation:</u>
For the given chemical reaction:
The expression of for above reaction follows:
........(1)
We are given:
Putting values in above equation, we get:
To calculate the molarity of solution, we use the equation:
Moles of hydrogen gas = 0.30 mol
Volume of solution = 2.0 L
Putting values in above equation, we get:
When hydrogen gas is added, the concentration of product gets increased. But, by Le-Chatelier's principle, the equilibrium will shift in the direction where concentration of product must decrease, which is in the backward direction.
Concentration of hydrogen gas when equilibrium is re-established = 0.1 + 0.15 = 0.25 M
Now, the equilibrium is shifting to the reactant side. The equation follows:
Initial: 0.1 0.1 0.4 0.1
At eqllm: 0.1+x 0.1+x 0.4-x 0.25-x
Putting values in expression 1, we get:
Neglecting the negative value of 'x'
Calculating the concentrations of the species:
Concentration of carbon dioxide = (0.4 - x) = (0.4 - 0.038) = 0.362 M
Concentration of hydrogen gas = (0.25 - x) = (0.25 - 0.038) = 0.212 M
Concentration of carbon monoxide = (0.1 + x) = (0.1 + 0.038) = 0.138 M
Concentration of water = (0.1 + x) = (0.1 + 0.038) = 0.138 M
Hence, the concentration of carbon dioxide, hydrogen gas, carbon monoxide and water when equilibrium is re-established are 0.362 M, 0.212 M, 0.138 M and 0.138 M respectively.