Destroyer G-chem Question

navneetdh

I have teh old version 2008 and my question number is 48

They gave the rxn at Equib as follows

2A (g) + B (g) <====> 2C (g)

0.50 atm of A and 0.20 atm of B are placed in a flask at 300K. At Equib the total pressure is found to be 0.60atm. Calculate the Kp for the reaction.

A. 15.0
B. 4.4
C. 6.6
D. 8.1
E. 1.0 x 10^-4

I dont understand why they used the ICE method.

I did it by assuming that mole fraction of A is 2/5 and calculating the PP by multiplying 2/4 with 0.6
then for B by calculation mole fraction of B is 1/5 and multiplying that to 0.60
finall mole fraction of C is 2/5 and multiplying that to 0.6

then putting each of these partial pressures in the Equation

K = ( [C]^ 2/[A]^2 )

Why am I wrong??

Shinpe

0.5 and 0.2 atm are the partial pressures before the reaction starts. Once the reaction gets to equilibrium, the A and B PPs are not gonna be 0.5 and 0.2 anymore. The solution shows how to get the equilibrium concentrations, which u would have to use to find K. I don't know what ICE means.

OP
N

navneetdh

0.5 and 0.2 atm are the partial pressures before the reaction starts. Once the reaction gets to equilibrium, the A and B PPs are not gonna be 0.5 and 0.2 anymore. The solution shows how to get the equilibrium concentrations, which u would have to use to find K. I don't know what ICE means.
ICE is Initial Change and Equilibrium.

Hmm I was thinking more like when you mix the gases up then their partial pressures in the system are going to be equal to their mole fractions times the total pressure in the container. Daltons Law.

So I just skipped to calculate the PP of each gas at Equib from the total pressure at equib.

Does that not work here because there is a reaction going on maybe??

Shinpe

ICE is Initial Change and Equilibrium.

Hmm I was thinking more like when you mix the gases up then their partial pressures in the system are going to be equal to their mole fractions times the total pressure in the container. Daltons Law.

So I just skipped to calculate the PP of each gas at Equib from the total pressure at equib.

Does that not work here because there is a reaction going on maybe??
2A (g) + B (g) <====> 2C (g)
is saying that 2 moles of A react with 1 mole of B to get you 2 moles of C. That is not necessarily the ratio of moles in the final equilibrium.
As is not here, you'd get 0.3 atm A, 0.1 atm B, and 0.2 atm C. The mole ratio is 3:1:2, not 2:1:2 as you'd guess from the equation.

OP
N

navneetdh

2A (g) + B (g) <====> 2C (g)
is saying that 2 moles of A react with 1 mole of B to get you 2 moles of C. That is not necessarily the ratio of moles in the final equilibrium.
As is not here, you'd get 0.3 atm A, 0.1 atm B, and 0.2 atm C. The mole ratio is 3:1:2, not 2:1:2 as you'd guess from the equation.
oooooooo  OMG thanks I feel like slapping myself for not realizing that those moles are not the total moles in the stupid container and therefore cant be used for mole ratios

sixkiller

I have teh old version 2008 and my question number is 48

They gave the rxn at Equib as follows

2A (g) + B (g) <====> 2C (g)

0.50 atm of A and 0.20 atm of B are placed in a flask at 300K. At Equib the total pressure is found to be 0.60atm. Calculate the Kp for the reaction.

A. 15.0
B. 4.4
C. 6.6
D. 8.1
E. 1.0 x 10^-4

I dont understand why they used the ICE method.

I did it by assuming that mole fraction of A is 2/5 and calculating the PP by multiplying 2/4 with 0.6
then for B by calculation mole fraction of B is 1/5 and multiplying that to 0.60
finall mole fraction of C is 2/5 and multiplying that to 0.6

then putting each of these partial pressures in the Equation

K = ( [C]^ 2/[A]^2 )

Why am I wrong??

You got ahead of yourself by assuming the mole fractions of the FINAL equilibrium state.

W/o making any assumptions with respect to the equilibrium state...we simply have the partial pressures in algebraic form.

2x, 0.5-2x, and 0.2-x respectively...

The only way we can deduce the mole fractions is if we have the FINAL partial pressures.

We cannot determine the FINAL partial pressures unless we have either Kp or the final total pressure at equilibrium.

OP
N

navneetdh

yup I realized what I was doing wrong.

Thank juuuuu alllll 