The equilibrium constant, Kc, for the reaction A(g) + B(g) <-->AB(g) is 115 at 25 degrees C. If 1.580 x 10^(‐1) mol of A and 4.721 x 10^(‐5) mol of B are introduced into a 1.00 liter vessel at 25 degrees C, calculate the equilibrium concentration of all species. Show your work using an ICE table.
..........A + B ==> AB
I.....0.1580..4.721E-5....0
C........-x....-x.........x
E....0.1580-x..4.721E-5-x..x
Substitute the E line into the Kc expression and solve for x, then evaluate A and B.
To calculate the equilibrium concentrations of all species, we can use an ICE table.
Let's start by writing the balanced equation for the reaction:
A(g) + B(g) <--> AB(g)
Next, let's set up the ICE table:
A(g) + B(g) <--> AB(g)
Initial: 0.1580 0.04721 0
Change: -x -x +x
Equilibrium: 0.1580-x 0.04721-x x
Now, we can plug in the given value of Kc and the initial concentrations into the equation for the equilibrium constant:
Kc = [AB(g)] / ([A(g)] * [B(g)])
Rearranging the equation:
Kc = (x) / (0.1580-x) * (0.04721-x)
Substituting the values:
115 = (x) / (0.1580-x) * (0.04721-x)
Now, we can solve this equation for x:
115(0.1580-x)(0.04721-x) = x
After solving this equation, we will get the value of x. Once we have that, we can substitute it back into the equilibrium expressions to find the equilibrium concentrations of A, B, and AB:
[A(g)] = 0.1580 - x
[B(g)] = 0.04721 - x
[AB(g)] = x
With these values, we have the equilibrium concentrations of all species.
To calculate the equilibrium concentration of all species, we can use an ICE table. ICE stands for Initial, Change, and Equilibrium concentrations. The steps involved are as follows:
Step 1: Write the balanced equation for the reaction.
A(g) + B(g) <--> AB(g)
Step 2: Set up the ICE table.
A(g) + B(g) <--> AB(g)
Initial 1.580 x 10^(-1) 4.721 x 10^(-5) 0
Change -x -x +x
Equilibrium 1.580 x 10^(-1) - x 4.721 x 10^(-5) - x x
Step 3: Set up the expression for the equilibrium constant.
Kc = [AB(g)] / [A(g)] * [B(g)]
For this reaction, the equilibrium constant (Kc) is given as 115.
115 = (1.580 x 10^(-1) - x) * (4.721 x 10^(-5) - x) / x
Step 4: Solve for x.
Multiply both sides by x:
115x = (1.580 x 10^(-1) - x) * (4.721 x 10^(-5) - x)
Expand the right side:
115x = 7.444 x 10^(-6) + x - 6.031 x 10^(-11) - x^2
Rearrange the equation:
x^2 - (7.444 x 10^(-6) - x - 6.031 x 10^(-11)) = 0
Simplify:
x^2 - 7.444 x 10^(-6) + x + 6.031 x 10^(-11) = 0
Now, solve this quadratic equation for x. You can use the quadratic formula or another method of solving quadratic equations.
Step 5: Once you have calculated the value of x, you can substitute it back into the ICE table to find the equilibrium concentrations of all species.
[A(g)] = 1.580 x 10^(-1) - x
[B(g)] = 4.721 x 10^(-5) - x
[AB(g)] = x
Note: The units for equilibrium concentrations will depend on the units used for the initial amounts of reactants and the volume of the vessel. In this case, the units are in moles per liter.
Remember to check that the value of x is small compared to the initial concentrations to ensure that the approximation made in the ICE table is valid.