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?
What is the Kp at 1439 °C for the reaction:
2CO(g) + O2(g) 2CO2(g)
If Kc is 2.5 x 1010 at the same temperature
a. repeat from the previous post.
b.
Kp = Kc*(RT)^delta n
To find the concentration of CO in the first question, we can use the equilibrium constant expression and the known values of moles and volume.
The equilibrium constant expression for the given equation is:
Kc = [CH3OH] / ([H2]^2 * [CO])
Here, [CH3OH] represents the concentration of CH3OH, [H2] represents the concentration of H2, and [CO] represents the concentration of CO.
We know the equilibrium constant, which is given as 19.
We also know the moles of CH3OH, which is given as 5.79 x 10^-3 moles.
The moles of H2 is given as 3.11 x 10^-2 moles.
The volume of the flask is given as 6.75 L.
To find the concentration of CO, we need to determine the initial concentrations of CH3OH and H2.
To do this, we can divide the moles of CH3OH and H2 by the volume of the flask:
[CH3OH] = (5.79 x 10^-3 moles) / (6.75 L)
[H2] = (3.11 x 10^-2 moles) / (6.75 L)
Now, we can substitute these values into the equilibrium constant expression:
19 = (5.79 x 10^-3 moles) / (6.75 L) / ([3.11 x 10^-2 moles / (6.75 L)]^2 * [CO])
Simplifying the expression:
19 = (5.79 x 10^-3) / (3.11 x 10^-2 / 6.75 L)^2 * [CO]
To find [CO], we can rearrange this expression:
[CO] = (5.79 x 10^-3) / (3.11 x 10^-2 / 6.75 L)^2 / 19
Evaluating this expression will give us the concentration of CO.
In the second question, we are given the equilibrium constant (Kc) as 2.5 x 10^10 at a particular temperature. We need to find the equilibrium constant in terms of partial pressure (Kp) at 1439 °C.
To convert Kc to Kp, we can use the relationship:
Kp = Kc * (RT)^Δn
Where R is the gas constant (0.0821 L atm / mol K), T is the temperature in Kelvin, and Δn represents the change in the number of moles of gaseous products minus the number of moles of gaseous reactants.
In this case, the equation is:
2CO(g) + O2(g) <=> 2CO2(g)
The Δn can be calculated as (2 - 0) - (1 + 0) = 1
Substituting the values into the equation:
Kp = (2.5 x 10^10) * ([0.0821 L atm / mol K] * [1439 + 273.15] K)^1
Simplifying and evaluating the expression will give us the value of Kp at 1439 °C.