at 700 k the equilibrium constant for the reactions is Kp=0.76. Carbon tertrachrolide gas yields carbon solid and chlorine gas. A flask is charged with 2.00 atm of CCl4, which then reaches equilibrium at 700K. A) what is the fraction of the CCl4 is converted?
b) what are the partial pressures of CCl4 and Cl2 at equlibrium
........CCl4(g) ==> C(s) + 2Cl2(g)
I........2..........0........0
C.......-p..................+2p
E.......2-p.................+2p
Substitute into Kp expression and solve for p. 2p will give you pCl2 and 2-p will give you pCCl4.
For part a, fraction is pCl2/pCCl4
To solve these questions, we will need to use the ideal gas law and the equilibrium expression for the given reaction. Let's go step by step:
a) To find the fraction of CCl4 that is converted, we need to consider the stoichiometry of the reaction. The balanced equation for the reaction is:
CCl4 (g) -> C (s) + 2Cl2 (g)
Here, we see that for every 1 mole of CCl4 that reacts, we get 1 mole of C (solid) and 2 moles of Cl2. Since the equilibrium constant, Kp, is given as 0.76, we can use the equilibrium expression:
Kp = (P_C * P_Cl2^2) / P_CCl4
Since we have the value of Kp and the pressure of CCl4, we can rearrange the equation to solve for the fraction of CCl4 converted:
Kp = (P_C * P_Cl2^2) / P_CCl4
0.76 = (1 * P_Cl2^2) / 2.00
Now, we can solve for P_Cl2:
0.76 * 2.00 = P_Cl2^2
1.52 = P_Cl2^2
Taking the square root of both sides, we get:
P_Cl2 = sqrt(1.52)
P_Cl2 ≈ 1.23 atm
Now, we can find the fraction of CCl4 converted:
Fraction of CCl4 converted = (initial CCl4 - equilibrium CCl4) / initial CCl4
= (2.00 - P_Cl2) / 2.00
= (2.00 - 1.23) / 2.00
≈ 0.385
Therefore, approximately 38.5% of the CCl4 is converted at equilibrium.
b) To find the partial pressures of CCl4 and Cl2 at equilibrium, we already know the pressure of Cl2 from part (a), which is approximately 1.23 atm. To find the partial pressure of CCl4, we can subtract the pressure of Cl2 from the initial pressure of CCl4:
Partial pressure of CCl4 = Initial pressure of CCl4 - Pressure of Cl2
= 2.00 - 1.23
≈ 0.77 atm
Therefore, at equilibrium, the partial pressure of CCl4 is approximately 0.77 atm and the partial pressure of Cl2 is approximately 1.23 atm.