Given the reaction and information below, each of the values asked for below. (make an I.C.E. table)

Question

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.

Answer 1

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.

Answer 2

Thanks i managed to calculate kp as 0.3

Answer 3

Looks OK to me.