An equilibrium mixture of hydrogen, carbon dioxide, water and carbon monoxide is in a closed, 1 L container at a fixed temperature as shown: H2(g) + CO2(g) ⇌ H2O(g) + CO(g) Keq = 1.600 The initial concentrations are [H2] = 1.000 mol L^{-1}, [CO2] = 0.500 mol L^{-1}, [H2O] = 0.400 mol L^{-1}, [CO] = 2.000 mol L^{-1}. An unknown amount of CO(g) was added to the same container, and the temperature was kept constant. After the new equilibrium had been established, the concentration of H2O(g) was found to be 0.200 mol L^{-1}. Using this information, calculate the unknown amount (in mol) of CO(g) that was added to the container.

SSCE HSC Chemistry Alcohols: An equilibrium mixture of hydrog

An equilibrium mixture of hydrogen, carbon dioxide, water and carbon monoxide is in a closed, 1 L container at a fixed temperature as shown: H2(g) + CO2(g) ⇌ H2O(g) + CO(g) Keq = 1.600 The initial concentrations are [H2] = 1.000 mol L^{-1}, [CO2] = 0.500 mol L^{-1}, [H2O] = 0.400 mol L^{-1}, [CO] = 2.000 mol L^{-1} - HSC - SSCE Chemistry - Question 30 - 2024 - Paper 1

Question 30

$An-equilibrium-mixture-of-hydrogen,-carbon-dioxide,-water-and-carbon-monoxide-is-in-a-closed,-1-L-container-at-a-fixed-temperature-as-shown:--H2(g)-+-CO2(g)-⇌-H2O(g)-+-CO(g)--Keq-=-1.600--The-initial-concentrations-are---[H2]-=-1.000-mol-L^{-1},---[CO2]-=-0.500-mol-L^{-1},---[H2O]-=-0.400-mol-L^{-1},---[CO]-=-2.000-mol-L^{-1}-HSC-SSCE Chemistry-Question 30-2024-Paper 1.png$

An equilibrium mixture of hydrogen, carbon dioxide, water and carbon monoxide is in a closed, 1 L container at a fixed temperature as shown: H2(g) + CO2(g) ⇌ H2O(g)... show full transcript

Worked Solution & Example Answer:An equilibrium mixture of hydrogen, carbon dioxide, water and carbon monoxide is in a closed, 1 L container at a fixed temperature as shown: H2(g) + CO2(g) ⇌ H2O(g) + CO(g) Keq = 1.600 The initial concentrations are [H2] = 1.000 mol L^{-1}, [CO2] = 0.500 mol L^{-1}, [H2O] = 0.400 mol L^{-1}, [CO] = 2.000 mol L^{-1} - HSC - SSCE Chemistry - Question 30 - 2024 - Paper 1

Step 1

Calculate the Change in [H2O]

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Answer

The initial concentration of H2O is 0.400 mol L^{-1} and the equilibrium concentration is 0.200 mol L^{-1}. Therefore, the change in concentration can be calculated as:

ext{Change in } [H2O] = (0.400 - 0.200) = 0.200 ext{ mol L}^{-1}

Step 2

ICE Table for the Reaction

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Answer

To analyze the equilibrium, we set up an ICE (Initial, Change, Equilibrium) table:

Species	Initial (mol)	Change (mol)	Equilibrium (mol)
H2	1.000	+0.200	1.200
CO2	0.500	+0.200	0.700
H2O	0.400	-0.200	0.200
CO	2.000 + x	-0.200	1.800 + x

Where x is the unknown amount of CO added.

Step 3

Use K_eq to Find x

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Answer

We use the equilibrium constant expression:

K_{eq} = rac{[H2O][CO]}{[H2][CO2]} = 1.600

Substituting the equilibrium concentrations:

1.600 = rac{0.200(1.800 + x)}{1.200(0.700)}

Next, simplify and solve for x:

Multiply both sides by the denominator:

1.600 imes 1.200 imes 0.700 = 0.200(1.800 + x)

Resulting in:

1.344 = 0.200 imes 1.800 + 0.200x

Calculating:

1.344 = 0.360 + 0.200x

Rearranging gives:

0.200x = 1.344 - 0.360

Thus,

0.200x = 0.984

Finally, solving for x:

x = rac{0.984}{0.200} = 4.92 ext{ mol in 1 L}

Calculate the Change in [H2O]

ICE Table for the Reaction

Use K_eq to Find x

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