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Consider the following system in a 1.00 L container: A(g) + B(g) = 2C(g) The equilibrium concentrations at 200°C were determined to be: [A] = 0.200 M [B] = 3.00 M [C]=0.500 M How many moles of A must be added to increase the concentration of C to 0.700 M at 200°C?​

User JanDotNet
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We can start by using the equilibrium concentrations of A, B, and C to calculate the equilibrium constant (Kc) for the reaction:

Kc = [C]^2 / ([A] x [B])

Substituting the given concentrations, we get:

Kc = (0.500 M)^2 / (0.200 M x 3.00 M) = 0.4167

Next, we can use the equilibrium concentrations to write an ICE table (Initial, Change, Equilibrium) to determine how the concentrations will change when we add more A to the system:

A(g) + B(g) = 2C(g)

I 0.200 M 3.00 M 0 M

C -x -x 2x

E 0.200 M - x 3.00 M - x 0.500 M + 2x

We know that we want the concentration of C to be 0.700 M, so we can set up an equation to solve for x:

0.700 M = 0.500 M + 2x

0.200 M = 2x

x = 0.100 M

This tells us that we need to add 0.100 moles of A to the system to increase the concentration of C from 0.500 M to 0.700 M at 200°C. To calculate the number of moles of A required, we can use the initial volume of the system, which is 1.00 L, and the concentration of A:

moles of A = volume x concentration = 1.00 L x 0.200 M = 0.200 moles

Therefore, we need to add 0.100 moles of A to the system, which is half the amount of A present initially.

User Shaun Cockerill
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