Given the reaction and information below, each of the values asked for below. (make an I.C.E. table)
A(g)+2B(g)↔C(g)
You start with 2. atm of each of the reactants in a flask. After a sufficient amount of time the reaction reaches equilibrium at 400K. You measure the partial pressure of C at equilibrium to be 0.5 atm. Fill in the missing information from the I.C.E. "chart"
A B C
I 2 2 0
C +0.5
E 0.5
I know I for A and B are 2 and I for C is 0.5. I need help with the C for A and B.
You're ICE chart becasue of the lack of spacing, leaves much to be desired. I've followed the problem and not your ICE chart.
................A(g)+2B(g)↔C(g)
I................2.0......2.0........0
C...............-p.......-2p.......+p
E............2.0-p....2.0-2p....0.5
If C is 0.5 @ equilibrium, the 0 + p = 0.5; therefore, p is 0.5
You know in line E that 2.0-p = 2.0-0.5 = 1.5 for A @ equilibrium and for B at equilibrium you know 2.0-2p = 2.0-1.0 = 1.0 so the completed chart looks like this.
................A(g)+2B(g)↔C(g)
I................2.0......2.0........0
C...............-0.5.......-1.0...+0.5
E...............1.5..........1.0....0.5
So you can use the E line to calculate Kp.
Thanks i managed to calculate kp as 0.3
Looks OK to me.
To determine the change in concentration (C) for A and B, we can use the stoichiometry of the balanced equation. Since the reaction stoichiometry is 1:2:1 (A:B:C), for every 1 mole of A that reacts, 2 moles of B will react, and 1 mole of C will be formed.
Therefore, the change in concentration (C) for A will be -x (since it has been consumed), and the change in concentration (C) for B will be -2x (twice of the change in concentration for A).
Now, let's fill in the missing information in the I.C.E. table:
A B C
I 2 2 0
C -x -2x +0.5
E 2-x 2-2x 0.5
Since the initial concentration (I) of A and B is 2 atm, and their change in concentration (C) is -x and -2x, respectively, the equilibrium concentration (E) for A will be (2 - x) atm, and for B, it will be (2 - 2x) atm.
Please note that the equilibrium concentration of C is given as 0.5 atm, which is filled in the I.C.E. table.
To fill in the missing information for the change (C) in the I.C.E. (Initial-Change-Equilibrium) table, we can use the stoichiometry of the balanced chemical equation.
From the balanced equation: A(g) + 2B(g) ↔ C(g)
The stoichiometric coefficient for A is 1, which means that for every mole of A that reacts, 1 mole of C is produced. Therefore, at equilibrium, the change in the amount of A (C[A]) will be the same as the change in the amount of C (C[C]), but with opposite signs.
Similarly, the stoichiometric coefficient for B is 2, which means that for every 2 moles of B that react, 1 mole of C is produced. Therefore, at equilibrium, the change in the amount of B (C[B]) will be half the change in the amount of C (C[C]), but with the same sign.
Given that the change in the amount of C is +0.5 atm, we can fill in the missing information for the change in the amount of A and B as follows:
C[A] = -0.5 atm (opposite sign to C[C])
C[B] = -0.25 atm (half the magnitude of C[C] with the same sign)
By subtracting the change values from the initial values, we can calculate the equilibrium values:
E[A] = I[A] + C[A] = 2 atm - 0.5 atm = 1.5 atm
E[B] = I[B] + C[B] = 2 atm - 0.25 atm = 1.75 atm
E[C] = I[C] + C[C] = 0.5 atm + 0.5 atm = 1 atm
Now the I.C.E. table is complete:
A B C
I 2 2 0
C -0.5 -0.25 +0.5
E 1.5 1.75 1
Note: The values in the table are in atm and represent the equilibrium partial pressures of each component.