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)

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 answer is the first option

explanation:

just googled it m8. not that hard

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the potential energy of the reactants is 200j.

from the energy diagram, the energy of the product formed is 350j; this means that, this reaction is an endothermic reaction, because it absorbs energy from its environment.