For the following reaction, the equilibrium constant Kc = 97.0 at 900K. If the initial concentrations of NH3 and H2S are both 0.20 M, what is the equilibrium concentration of H2S?
H2S(g) + NH3(g) = NH4HS(s)
Would you have to create an ICE chart for this type of problem?
I make an ICE chart for almost every chemistry problem.
Yes, for this type of problem, you would typically create an ICE (Initial, Change, Equilibrium) chart to determine the equilibrium concentration of H2S. An ICE chart is a systematic way to organize the information and track the changes that occur during a reaction.
Here's how to create an ICE chart for this problem:
1. Write down the balanced chemical equation:
H2S(g) + NH3(g) = NH4HS(s)
2. Identify the initial concentrations of NH3 and H2S:
Given that the initial concentrations of NH3 and H2S are both 0.20 M.
3. Set up the ICE chart:
NH3(g) + H2S(g) = NH4HS(s)
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I 0.20 0.20 0
The "I" row represents the initial conditions, where we write down the initial concentrations of NH3 and H2S.
4. Determine the change in concentration:
We'll let "x" represent the change in concentration of H2S. Since both NH3 and H2S have the same initial concentration and react in a 1:1 ratio according to the balanced equation, the change in concentration of H2S will be -x.
NH3(g) + H2S(g) = NH4HS(s)
----------------------------------
I 0.20 0.20 0
C -x -x
E 0.20-x 0.20-x 0
The "C" row represents the change in concentration, where we subtract the value of x from the initial concentrations of NH3 and H2S.
5. Determine the equilibrium concentrations:
We know that the equilibrium constant, Kc, is equal to the ratio of the concentrations of the products and reactants at equilibrium. In this case, Kc = 97.0.
Kc = [NH4HS] / [NH3] * [H2S]
We want to find the equilibrium concentration of H2S, so we'll fill in the values we know and solve for x:
97.0 = 0 / (0.20-x) * (0.20-x)
Simplifying the equation:
97.0 = (0.20-x)^2
Rearranging and taking the square root of both sides:
0.20-x = √97.0
Solving for x:
x = 0.20 - √97.0
6. Calculate the equilibrium concentration of H2S:
Now that we know the value of x, we can substitute it back into the expression for H2S in the ICE chart to find the equilibrium concentration:
[H2S]equilibrium = 0.20 - x
[H2S]equilibrium = 0.20 - (0.20 - √97.0)
[H2S]equilibrium = √97.0 M
Therefore, the equilibrium concentration of H2S is approximately equal to √97.0 M.