H2(g) + CO2(g)<-> H20(g) + CO(g)

When H2(g) is mixed with CO2(g) at 2,000K, equilibrium is achieved according to the equation above. In one experiment, the following equilibrium concentrations were measured:
[H2]=0.20 M
[CO2]=0.30 M
[H2O]=[CO]=0.55M
a) What is the mole fraction of CO(g) in the equilibrium mixture?
b)Using the equilibrium concentrations given above, calculate the value of Kc, the equilibrium constant for the reaction.
c) Determine Kp,in terms of Kc for this system.
d)When the system is cooled from 2,000 K to a lower temperature, 30.0% of the CO(g) is converted back to CO2(g). Calculate the value of Kc at this lower temperature.
e)In a different experiment, 0.50 mole of H2(g) is mixed with 0.50 mole of CO2(g) in a 3.0 L reaction vessel at 2,000K. Calculate the equilibrium concentration, in moles/liter, of CO(g) at this temperature.

I know this is lengthy and if i at least get a good start on this...maybe I will know how to do the rest...but please help me!! Thanks!

Oh my goodness... what grade are you in? That has got to be the most confusing problem I have ever seen in my life!! (Of course I'm only barely in middle school!)

H2(g) + CO2(g)<-> H20(g) + CO(g)
When H2(g) is mixed with CO2(g) at 2,000K, equilibrium is achieved according to the equation above. In one experiment, the following equilibrium concentrations were measured:
[H2]=0.20 M
[CO2]=0.30 M
[H2O]=[CO]=0.55M
a) What is the mole fraction of CO(g) in the equilibrium mixture?

This looks straight forward. XCO=(mols CO/mols CO + mols H2 + mols H2O + mols CO2)

b)Using the equilibrium concentrations given above, calculate the value of Kc, the equilibrium constant for the reaction.

This looks straight forward. Kc=(CO)(H2O)/(CO2)(H2).
Simply plug in the concentrations and calculate Kc.


c) Determine Kp,in terms of Kc for this system.

Look in your text/notes and find the conversion from Kc to Kp.

d)When the system is cooled from 2,000 K to a lower temperature, 30.0% of the CO(g) is converted back to CO2(g). Calculate the value of Kc at this lower temperature.

Subtract 30% CO to obtain the new CO concentration, add 30% to CO2 to obtain the new CO2 concentration. Same thing for H2O and H2 concentrations. Plug the new numbers into Kc expression above and calculate the new Kc.

e)In a different experiment, 0.50 mole of H2(g) is mixed with 0.50 mole of CO2(g) in a 3.0 L reaction vessel at 2,000K. Calculate the equilibrium concentration, in moles/liter, of CO(g) at this temperature.

Use Kc found in your earlier parts at 2000 K. Post your work for the other parts and we can help you get through this part. I suspect you will know how to do this part by the time you get here.

Check my thinking.
Check my work.

Simple, simple, simple...Just refer to your text book and it should all be there. Isn't calculating Kp so easy?!

(a) CO = f(0.55 mol, 1.6 mol) = 0.34

(b) Kc = ([H2O][CO])/([H2][CO2]) = (0.55×0.55)/(0.20×0.30) = 5.04
(c) since Δn = 0, Kc = Kp
(d) [CO] = 0.55 - 30.0% = 0.55 - 0.165 = 0.385 M, [H2O]= 0.55 - 0.165 = 0.385 M, [H2]= 0.20 + 0.165 = 0.365 M
[CO2] = 0.30 + 0.165 = 0.465 M K = (0.385)2/(0.365×0.465) = 0.87
(e) let X = Δ[H2] to reach equilibrium
[H2] = 0.50 mol/3.0L - X = 0.167 – X, [CO2] = 0.50 mol/3.0L - X = 0.167 – X, [CO] = +X ; [H2O] = +X
K = X2/(0.167 - X)2 = 5.04 ; X = [CO] = 0.12 M

a) To find the mole fraction of CO(g) in the equilibrium mixture, we need to divide the moles of CO by the total moles of all species. The mole fraction, represented by X, is calculated using the formula:

X_CO = moles of CO / (moles of CO + moles of H2 + moles of H2O + moles of CO2)

Here, we are given the equilibrium concentrations in Molarity, so we can directly use them to calculate the mole fractions. The moles of CO can be obtained by multiplying the concentration of CO by the volume of the equilibrium mixture. Since the volume is not given, we can assume an arbitrary volume, such as 1 L, for the sake of calculation:

moles of CO = [CO] * volume = 0.55 * 1 = 0.55 mol

The moles of H2, H2O, and CO2 can be calculated similarly:

moles of H2 = [H2] * volume = 0.20 * 1 = 0.20 mol
moles of H2O = [H2O] * volume = 0.55 * 1 = 0.55 mol
moles of CO2 = [CO2] * volume = 0.30 * 1 = 0.30 mol

Now we can plug these values into the mole fraction formula:

X_CO = 0.55 / (0.55 + 0.20 + 0.55 + 0.30) = 0.55 / 1.60 = 0.34375

Therefore, the mole fraction of CO(g) in the equilibrium mixture is 0.34375.

b) To calculate the equilibrium constant Kc, we use the formula:
Kc = ([H2O] * [CO]) / ([H2] * [CO2])

Given that [H2O] = [CO] = 0.55 M, and [H2] = 0.20 M, and [CO2] = 0.30 M, we can substitute these values into the equation:

Kc = (0.55 * 0.55) / (0.20 * 0.30) = 0.3025 / 0.06 = 5.0417

Therefore, the value of Kc, the equilibrium constant, is 5.0417.

c) To determine Kp in terms of Kc for this system, we need to use the appropriate conversion formula. In this case, since the balanced equation consists of gases, we can use the ideal gas law to relate Kp and Kc:

Kp = Kc * (RT)^Δn

Where Δn is the difference in the number of moles of gaseous products and reactants. In this equation, we can see that Δn is zero since the number of moles of gaseous products (H2O and CO) is equal to the number of moles of gaseous reactants (H2 and CO2).

Therefore, Kp = Kc * (RT)^0 = Kc * 1 = Kc

So, Kp is equal to Kc in this system.

d) To calculate the new value of Kc at the lower temperature where 30% of CO(g) is converted back to CO2(g), we need to adjust the concentrations accordingly. Since 30% of CO(g) is converted back to CO2(g), we subtract 30% from the current concentration of CO and add 30% to the current concentration of CO2:

[CO] = 0.55 - 0.30 * 0.55 = 0.55 - 0.165 = 0.385 M
[CO2] = 0.30 + 0.30 * 0.55 = 0.30 + 0.165 = 0.465 M

Substituting these new concentrations into the equilibrium constant expression:

Kc = ([H2O] * [CO]) / ([H2] * [CO2]) = (0.55 * 0.385) / (0.20 * 0.465) = 0.21175 / 0.093 = 2.2785

Therefore, the value of Kc at the lower temperature is 2.2785.

e) To calculate the equilibrium concentration of CO(g) in moles/liter at this temperature, we need to use the equilibrium constant Kc. Given that 0.50 mole of H2(g) and 0.50 mole of CO2(g) are mixed in a 3.0 L reaction vessel, we can calculate the initial concentrations of H2 and CO2:

[H2] = (moles of H2) / volume = 0.50 mol / 3.0 L = 0.1667 M
[CO2] = (moles of CO2) / volume = 0.50 mol / 3.0 L = 0.1667 M

Using the values of [H2], [CO2], and Kc from part b, we can rearrange the equation to solve for [CO]:

Kc = ([H2O] * [CO]) / ([H2] * [CO2])
[CO] = (Kc * [H2] * [CO2]) / [H2O]

Substituting the given values:

[CO] = (5.0417 * 0.1667 * 0.1667) / 0.55 = 0.0278 M

Therefore, the equilibrium concentration of CO(g) at this temperature is 0.0278 moles/liter.

I apologize for the confusion in my previous response. Let's break down the problem step by step:

a) To find the mole fraction of CO(g) in the equilibrium mixture, we need to calculate the moles of each component based on their concentrations. The mole fraction of a component is given by the formula:

Mole Fraction = moles of component / total moles of all components

Given the equilibrium concentrations:
[H2] = 0.20 M,
[CO2] = 0.30 M,
[H2O] = [CO] = 0.55 M,

To find the moles of CO, we multiply its concentration by the volume of the system. Since the volumes are not given, we assume they are equal and cancel out in the ratio. Therefore:
moles of CO = [CO] * volume = 0.55 M * 1 L (assuming 1 L system volume) = 0.55 moles

Similarly, we find the moles of other components:
moles of H2 = [H2] * volume = 0.20 M * 1 L = 0.20 moles
moles of CO2 = [CO2] * volume = 0.30 M * 1 L = 0.30 moles
moles of H2O = [H2O] * volume = 0.55 M * 1 L = 0.55 moles

Now, we can calculate the total moles of all components:
total moles = moles of CO + moles of H2 + moles of H2O + moles of CO2
total moles = 0.55 moles + 0.20 moles + 0.55 moles + 0.30 moles
total moles = 1.60 moles

Finally, we can find the mole fraction of CO:
Mole Fraction of CO = moles of CO / total moles of all components
Mole Fraction of CO = 0.55 moles / 1.60 moles = 0.3437

Therefore, the mole fraction of CO in the equilibrium mixture is approximately 0.3437.

b) To calculate the value of Kc, the equilibrium constant for the reaction, we use the formula:

Kc = ([H2O] * [CO]) / ([H2] * [CO2])

Plugging in the given equilibrium concentrations:
Kc = (0.55 M * 0.55 M) / (0.20 M * 0.30 M)
Kc = 1.210 / 0.06
Kc = 20.17

Therefore, the value of Kc for the reaction is approximately 20.17.

c) To determine Kp in terms of Kc for this system, you need to refer to your textbook or notes for the specific equation. The conversion from Kc to Kp depends on the stoichiometry of the reaction and the ideal gas law. Please consult your resources for the correct equation.

d) To calculate the new value of Kc at a lower temperature, where 30.0% of the CO(g) is converted back to CO2(g), we need to adjust the concentrations accordingly.

Given that 30.0% of the CO converts to CO2, the new concentration of CO can be calculated as:
new [CO] = original [CO] * (1 - 0.30)
new [CO] = 0.55 M * 0.70
new [CO] = 0.385 M

Similarly, the new concentration of CO2 is:
new [CO2] = original [CO2] + (original [CO] * 0.30)
new [CO2] = 0.30 M + (0.55 M * 0.30)
new [CO2] = 0.30 M + 0.165 M
new [CO2] = 0.465 M

Using the same formula as in part (b), calculate the new value of Kc with the adjusted concentrations of CO and CO2.

e) To calculate the equilibrium concentration of CO(g) at a given temperature, you can use the mole ratios from the balanced equation and the given number of moles for H2 and CO2.

From the balanced equation, the stoichiometry is 1:1 for H2 and CO2, and 1:1 for H2O and CO. This means that for every mole of H2 used, we will have one mole of CO produced.

Given 0.50 moles of H2 and 0.50 moles of CO2, we know that 0.50 moles of CO will be produced.

Since we have a total volume of 3.0 L, the equilibrium concentration can be calculated as:
[CO] = moles CO / volume
[CO] = 0.50 moles / 3.0 L
[CO] = 0.167 M

Therefore, the equilibrium concentration of CO(g) at this temperature is approximately 0.167 M.

I hope this helps you in solving the problem. If you need further clarification or assistance, please let me know.