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A mixture of 0.500 mole of carbon monoxide and 0.400 mole of bromine was placed into a rigid 1.00-L container and the system was allowed to come to equilibrium. The equilibrium concentration of COBr2 was 0.233 M. What is the value of Kc for this reaction? CO(g) + Br2(g) COBr2(g)

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Answer:

The value for the equilibrium constant Kc is 5.23

Explanation:

Step 1: Data given

Number of moles CO = 0.500 moles

Number of moles Br2 = 0.400 moles

Volume = 1.0 L

The equilibrium concentration of COBr2 was 0.233 M.

Step 2: The balanced equation

CO(g) + Br2(g) ⇆ COBr2(g)

Step 3: The initial concentrations

Concentration = mol /volume

[CO] = 0.500 moles / 1L = 0.500 M

[Br2] = 0.400 moles / 1L = 0.400 M

[COBr2] = 0M

Step 4: The concentration at the equilibrium

[CO] = 0.500 - X M

[Br2] =  0.400 - X M

[COBr2] = XM = 0.233 M

[CO] = 0.500 - 0.233 =  0.267 M

[Br2] =  0.400 - 0.233 = 0.167 M

[COBr2] = XM = 0.233 M

Step 5: Calculate Kc

Kc = [COBr2]/[CO][Br2]

Kc = 0.233 / (0.267*0.167)

Kc = 5.23

The value for the equilibrium constant Kc is 5.23

The value for the equilibrium constant Kc is 5.23

Calculation of the value for the equilibrium constant:

Since A mixture of 0.500 mole of carbon monoxide and 0.400 mole of bromine was placed into a rigid 1.00-L container. And, The equilibrium concentration of COBr2 was 0.233 M.

Now The initial concentrations

Concentration = mol /volume

For [CO] = 0.500 moles / 1L = 0.500 M

For [Br2] = 0.400 moles / 1L = 0.400 M

For [COBr2] = 0M

Now The concentration at the equilibrium

For [CO] = 0.500 - X M

For [Br2] =  0.400 - X M

For [COBr2] = XM = 0.233 M

Now

[CO] = 0.500 - 0.233 =  0.267 M

[Br2] =  0.400 - 0.233 = 0.167 M

[COBr2] = XM = 0.233 M

Now  Kc

Kc = [COBr2]/[CO][Br2]

Kc = 0.233 / (0.267*0.167)

Kc = 5.23

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