[Chem] Ideal Gases

[Nava2]

Carbon Monoxide reacts with hydrogen gas to form methane and water.

CO + 3H2 -> CH4 + H2O(g)
A mixture initially contains CO and excess H2, and the total pressure of this mixture is 65.0kPa. After the reaction is complete, the pressure has fallen to 35.0 kPa. What was the mole fraction of H2 in the initial mixture? (Assume constant volume and constant temperature.)

A) 0.462
B) 0.692
C) 0.538
D) 0.769
E) 0.923

----------------------

This question was on my exam, I have NO idea how to solve it.

[tarajunky]

You've got 4 moles of gas (plus some extra H2) being converted to 2 moles of gas (plus less H2), so the pressure drops.

Since there is excess H2, that would mean that the initial molar fraction would have to be greater than 3/4. That narrows it down to D or E. (If the molar fraction was less than .75 then there would be excess CO.)

My first guess would be D, so I would try plugging in that value to see what happens.

Let's assume that 1 Mole of gas causes 65 kPa pressure.

.231M CO + .769M H2 -> .231M CH4 + .231M H20 + .076M H2 ( .769 - .231*3 = 0.076)

1M ~ 65 kPa
.231M + .231M + .076M = 0.538M * 65 kPa/1M = 34.97 kPa

[Nava2]

I guessed A, but I really don't see what you did.

I don't think it should need to be guessed like that either. There SHOULD be an answer.

By the way, I really appreciate the help. :)

[mastaraymond]

pV = nRT

[tarajunky]

D IS the correct answer, which is what my math showed.

I haven't taken a chemistry class in 10 years, so I just used my own foggy understanding to figure it out.

First of all, the answer MUST be greater than 0.75. In the reaction, there are 3 moles H2 consumed for every 1 mole CO consumed. So, if there is no excess present, the molar ratio of H2 would be 3/4, or 0.75. (The molar ratio of CO would be 1/4, or 0.25) If there is excess H2, then its molar ratio must be greater than that. If there is excess CO, then the molar ratio of H2 would be less than 0.75.

To solve it with a generic formula you would just have to do something like this...

Moles CO = X
Moles H2 = 1-X

Then write the balanced equation.

(x + 1-x)/65 -> (x + x + (1-x) - 3x)) /35 // Write the ratio of Moles to Pressure for each.
35*(1) -> 65*(1 - 2x)
35 -> 65 - 130x
30/130 =x
x=0.2307
1-x = 0.7692

[Nava2]

D IS the correct answer, which is what my math showed.

I haven't taken a chemistry class in 10 years, so I just used my own foggy understanding to figure it out.

First of all, the answer MUST be greater than 0.75. In the reaction, there are 3 moles H2 consumed for every 1 mole CO consumed. So, if there is no excess present, the molar ratio of H2 would be 3/4, or 0.75. (The molar ratio of CO would be 1/4, or 0.25) If there is excess H2, then its molar ratio must be greater than that. If there is excess CO, then the molar ratio of H2 would be less than 0.75.

To solve it with a generic formula you would just have to do something like this...

Moles CO = X
Moles H2 = 1-X

Then write the balanced equation.

(x + 1-x)/65 -> (x + x + (1-x) - 3x)) /35 // Write the ratio of Moles to Pressure for each.
35*(1) -> 65*(1 - 2x)
35 -> 65 - 130x
30/130 =x
x=0.2307
1-x = 0.7692

*sigh* I tried to do something along those lines but was caught up with the fact that the excess is on both sides and I couldn't solve for it.

Thanks Tara :)