Calculate the value of Kp for the equation
C(s)+CO2(g)<->2CO(g) Kp=?
given that at a certain temperature
C(s)+2H2O(g)<->CO2(g)+2H2(g) Kp1=3.33
H2(g)+CO2(g)<->H2O(g)+CO(g) Kp2=.733
Please show detailed steps.
C(s)+2H2O(g)<->CO2(g)+2H2(g) Kp1=3.33
H2(g)+CO2(g)<->H2O(g)+CO(g) Kp2=.733
Multiply equation 2 by 2 and add to equation 1 to obtain:
C(s) + 2H2O + 2H2 + 2CO2 ==> CO2 + 2H2 + 2H2O + 2CO
Notice 2H2O cancels, 2H2 cancels, 1 CO2 cancels to leave
C(s) + CO2 ==> 2CO
Kp for eqn 1 is Kp1 as listed. Kp for equation 2 multiplied by 2 is K^2p and is K'p2 (0.733)^2. That's because when you multiply and equation by a coefficient(n) the k values is k^n.
When we ADD equations (as we did above), the combined k value is multiplied; i.e., kp1 x (Kp2)^2 = ?
Thank you so much for the detailed steps. Finally understood what I need to do.
Well, well, well, looks like we've got ourselves a stoichiometry party going on here! Let's break it down step by step:
We have three equations with their respective Kp values:
1. C(s) + 2H2O(g) <-> CO2(g) + 2H2(g) with Kp1 = 3.33
2. H2(g) + CO2(g) <-> H2O(g) + CO(g) with Kp2 = 0.733
3. C(s) + CO2(g) <-> 2CO(g) with Kp = ?
To get the value of Kp for the third equation, we'll use a bit of a trick. Since the first and second equations have CO2(g) as a common species, we can combine them in such a way that it gives us the third equation with the desired Kp.
1. C(s) + 2H2O(g) <-> CO2(g) + 2H2(g) (Kp1 = 3.33)
2. H2(g) + CO2(g) <-> H2O(g) + CO(g) (Kp2 = 0.733)
To get the desired reaction, we can multiply equation 1 by 2 and equation 2 by 1:
2 x (C(s) + 2H2O(g) <-> CO2(g) + 2H2(g)) gives us 2C(s) + 4H2O(g) <-> 2CO2(g) + 4H2(g)
1 x (H2(g) + CO2(g) <-> H2O(g) + CO(g)) gives us H2(g) + CO2(g) <-> H2O(g) + CO(g)
Now, let's add these two equations together:
2C(s) + 4H2O(g) + H2(g) + CO2(g) <-> 2CO2(g) + 4H2(g) + H2O(g) + CO(g)
If we arrange it a bit, we get:
2C(s) + 4H2(g) <-> 2CO(g) + 3H2O(g) + CO(g)
Notice that in this combined equation, we have the desired reaction:
C(s) + CO2(g) <-> 2CO(g)
Now let's look at the new equation we got after adding the two initial equations:
2C(s) + 4H2(g) <-> 2CO(g) + 3H2O(g) + CO(g)
The Kp value for this equation will be the product of Kp1 and Kp2, since we combined those equations to get this one. So we have:
Kp = Kp1 x Kp2
Kp = 3.33 x 0.733
And there you have it, the value of Kp for the equation C(s) + CO2(g) <-> 2CO(g) is approximately 2.44.
I hope that clowning around with stoichiometry brought a smile to your face!
To calculate the value of Kp for the equation C(s) + CO2(g) <-> 2CO(g), we can use the information given for Kp1 and Kp2.
The overall reaction is the sum of the two reactions:
C(s) + 2H2O(g) <-> CO2(g) + 2H2(g) (1)
H2(g) + CO2(g) <-> H2O(g) + CO(g) (2)
Since we have Kp1 and Kp2, we can use the fact that the equilibrium constant for a reaction in the reverse direction is the reciprocal of the equilibrium constant for the original reaction.
Recall that for the reaction C(s) + 2H2O(g) <-> CO2(g) + 2H2(g), Kp1 = 3.33.
(1) Reversing the reaction to obtain the reaction C(s) + 2H2O(g) <-> CO2(g) + 2H2(g), the equilibrium constant is the reciprocal of Kp1.
Kp_reverse = 1 / Kp1 = 1 / 3.33
Similarly, for the reaction H2(g) + CO2(g) <-> H2O(g) + CO(g), Kp2 = 0.733.
(2) Reversing the reaction to obtain the reaction CO(g) + H2O(g) <-> H2(g) + CO2(g), the equilibrium constant is the reciprocal of Kp2.
Kp_reverse2 = 1 / Kp2 = 1 / 0.733
Now, we can multiply the equations (1) and (2) together to obtain the desired reaction:
(C(s) + 2H2O(g) <-> CO2(g) + 2H2(g)) * (H2(g) + CO2(g) <-> H2O(g) + CO(g))
Since H2O(g) and CO2(g) appear on both sides of the reaction, they cancel out:
(C(s) + 2H2O(g) <-> CO2(g) + 2H2(g)) * (H2(g) + CO2(g) <-> H2O(g) + CO(g))
= (C(s) + 2H2(g)) * (CO(g))
= C(s)CO(g) + 2H2(g)CO(g)
Now, we can combine the equilibrium constants for the two reactions:
Kp_combined = Kp_reverse * Kp_reverse2
Substituting the values obtained earlier:
Kp_combined = (1 / 3.33) * (1 / 0.733)
Finally, simplify the expression to find Kp_combined:
Kp_combined ≈ 0.396
Therefore, the value of Kp for the equation C(s) + CO2(g) <-> 2CO(g) is approximately 0.396.
To calculate the value of Kp for the equation C(s) + CO2(g) ↔ 2CO(g), we can use the concept of equilibrium constants, which involves using the given equilibrium constants (Kp1 and Kp2) and manipulating them to find Kp for the desired reaction.
The equation you provided involving H2(g), CO2(g), H2O(g), and CO(g) can be rearranged as follows:
CO(g) + H2O(g) ↔ H2(g) + CO2(g)
To determine the value of Kp for this equation, we need to consider the individual equilibrium constants and their relationship to each other.
The equation you provided:
C(s) + 2H2O(g) ↔ CO2(g) + 2H2(g) (Equation 1) - Kp1 = 3.33
And the equation given:
H2(g) + CO2(g) ↔ H2O(g) + CO(g) (Equation 2) - Kp2 = 0.733
Let's now consider the desired equation:
C(s) + CO2(g) ↔ 2CO(g) (Equation 3) - Kp = ?
We can manipulate Equations 1 and 2 to obtain Equation 3.
Multiply Equation 1 by 2, and multiply Equation 2 by 2:
2(C(s) + 2H2O(g)) ↔ 2(CO2(g) + 2H2(g)) (Equation 4)
2(CO(g) + H2O(g)) ↔ 2(H2(g) + CO2(g)) (Equation 5)
Now let's combine Equations 4 and 5:
2(C(s) + 2H2O(g)) + 2(CO(g) + H2O(g)) ↔
2(CO2(g) + 2H2(g)) + 2(H2(g) + CO2(g))
Simplifying the equation, we have:
2C(s) + 6H2O(g) + 2CO(g) ↔ 4CO2(g) + 4H2(g)
Now we can compare the coefficients of Equation 3 with the simplified equation above to determine the relationship between their equilibrium constants (Kp):
2CO(g), 4CO2(g), 4H2(g)
Compared to the desired equation:
C(s) + CO2(g), 2CO(g)
We can see that the coefficients of CO(g) and CO2(g) in the simplified equation are twice the coefficients in the desired equation. Thus, the equilibrium constant Kp for the desired equation (Equation 3) is the square root of the product of the equilibrium constants of the simplified equation (Equation 4) and the given equation (Equation 2).
Kp = sqrt(Kp1 * Kp2)
= sqrt(3.33 * 0.733)
≈ 1.066
Therefore, the value of Kp for the equation C(s) + CO2(g) ↔ 2CO(g) is approximately 1.066.