The equilibrium constant for the equation

2 H2(g) + CO(g) CH3OH(g)
Is 19 at a certain temperature. If there are 3.11 x 10-2 moles of H2 and 5.79 x 10-3 moles of CH3OH at equilibrium in a 6.75 L flask. What is the concentration of CO?

At 1280 °C the equilibrium constant for the reaction
Br2(g) 2Br(g)
Is 1.1 x 10-3. If the initial concentrations are [Br2(g)] = 0.0310 M and [Br(g)] = 0.0210 M, calculate the concentrations of these 2 species in equilibrium

a.

Calculate (H2)&(CH3OH) from M = mols/L.
Then substitute these values into K expression and solve for the missing CO in M

b. I assume the K of 1.1E-3 is Kc and not Kp.
.........Br2(g) ==> 2Br(g)
I.....0.0310.......0.210
C........-x..........+2x
E....0.0310-x.......0.210+2x
Substitute the E line into the Kc expression and solve for x, then evaluate the other values.

To find the concentration of CO in the first equation, we can use the equilibrium constant expression:

K = [CH3OH] / ([H2]^2 * [CO])

We know that the equilibrium constant (K) is 19, and the concentrations of CH3OH and H2 are given. Let's say the concentration of CO is x.

Plug in the given values into the equation:

19 = (5.79 x 10^(-3)) / ((3.11 x 10^(-2))^2 * x)

Simplify the equation:

19 = (5.79 x 10^(-3)) / (9.6481 x 10^(-4) * x)

Cross-multiply:

19 * 9.6481 x 10^(-4) * x = 5.79 x 10^(-3)

Solve for x:

x = (5.79 x 10^(-3)) / (19 * 9.6481 x 10^(-4))

Calculate x:

x ≈ 0.310 M

Therefore, the concentration of CO in the equilibrium mixture is approximately 0.310 M.

Now let's move on to the second equation. The equilibrium constant expression is:

K = [Br(g)]^2 / [Br2(g)]

We know that the equilibrium constant (K) is 1.1 x 10^(-3), and the initial concentrations of Br2 and Br are given. Let's say the concentrations of Br2 and Br in equilibrium are x and y, respectively.

Plug in the given values into the equation:

1.1 x 10^(-3) = (y^2) / (0.0310 - x)

Simplify the equation:

1.1 x 10^(-3) = (y^2) / (0.0310 - x)

Cross-multiply:

1.1 x 10^(-3) * (0.0310 - x) = y^2

Expand the equation:

1.1 x 10^(-3) * 0.0310 - 1.1 x 10^(-3) * x = y^2

Solve for y:

y = sqrt(1.1 x 10^(-3) * 0.0310 - 1.1 x 10^(-3) * x)

Calculate y:

y ≈ 0.00423 M

Now, substitute the value of y back into the equation and solve for x:

0.00423 = sqrt(1.1 x 10^(-3) * 0.0310 - 1.1 x 10^(-3) * x)

Square both sides of the equation:

0.00423^2 = 1.1 x 10^(-3) * 0.0310 - 1.1 x 10^(-3) * x

Solve for x:

x = (1.1 x 10^(-3) * 0.0310 - 0.00423^2) / (1.1 x 10^(-3))

Calculate x:

x ≈ 0.0279 M

Therefore, in the equilibrium mixture, the concentration of Br2 is approximately 0.0279 M, and the concentration of Br is approximately 0.00423 M.