Gen Chem - Gases

[optomhopefully]

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Hey guys,

I'm working on this gen chem question & get as far as determining that H2 is the limiting reactant but them am not sure where to go from there.

Carbon monoxide, CO, and hydrogen, H2, react according to the reaction below.
2 CO(g) + 5 H2(g) --> C2H6(g) + 2 H2O(g)

What volume of the excess reactant remains if 20.4 L CO and 43.5 L H2 are allowed to react. Assume that the volumes of both gases are measured at 727oC and 1.19 atm.

Thanks for any help!

 

[Cool Beans]

For gases the volume of the gas is proportional to the moles of gas and so you can view this problem just as if you'd had 20.4moles CO and 43.5moles H2. From here it's just like any other limiting reagent problem. If you divide the number of moles by each reactants coefficient you can see how many times you can stoichiometrically perform the reaction with that amount of reactant:

20.4mol CO/2 = 10.2 times
43.5mol H2/5 = 8.7 times

So H2 is the limiting reagent and CO is the reagent in excess and we'll run out of H2 after performing the reaction as written 8.7 times. If we do this reaction 8.7 times, we'll use up 17.4 moles of CO and would therefore have 3 moles left over (in excess).

I've worked the problem out in moles cause I thought it would be easier to follow but it would have worked out the same way if we'd used the volumes instead and so you have 3L of CO in excess.

Hope this helps!

 

[optomhopefully]

Hey CoolBeans! Thanks for the help I'm just not sure what u mean by 8.7 times..do you mean 8.7 moles? The rest I was able to follow!

I had another 2 questions that was giving me some trouble if anyone has any idea:

1) A 1.7-L flask of H2(g) at 2.7 atm and a 3.2-L flask of He(g) at 2.4 atm are connected by a tube of negligible volume. (See the figure below.) A valve on the tube is opened, allowing the gases to mix at constant temperature with no reaction. What is the final pressure?

2) The enthalpy change for the reaction below at 25oC is -1,449 kJ (per mole of C7H16). What is the internal energy change for the reaction at 25oC?
7 CO(g) + 15 H2(g) --> C7H16(l) + 7 H2O(l)

Thanks for the help!!

 

[Cool Beans]

Hey CoolBeans! Thanks for the help I'm just not sure what u mean by 8.7 times..do you mean 8.7 moles? The rest I was able to follow!

What I meant was in regards to what the coefficients tell you. One way to look at it is that every time you perform the reaction, it requires 2mol CO and 5mol H2. So assuming I already knew H2 was the limiting reagent, if we had 5mol H2 we could perform the reaction 1 time; if we had 10mol H2 we could perform the reaction 2 times; if we had 40mol H2 we could perform the reaction 8 times; and finally if we had 43.5mol H2 we could perform the reaction 8.7 times (43.5/5). Let me know if this clears it up for you.

I had another 2 questions that was giving me some trouble if anyone has any idea:

1) A 1.7-L flask of H2(g) at 2.7 atm and a 3.2-L flask of He(g) at 2.4 atm are connected by a tube of negligible volume. (See the figure below.) A valve on the tube is opened, allowing the gases to mix at constant temperature with no reaction. What is the final pressure?

So what you'll want to do here is from the ideal gas law say that P = nRT/V. But you can't just add the 2 pressures together or even simply take an average. What you need to do to get the final pressure is plug in the final total volume (4.9L) and the final numer of moles of gas. It's tricky here because you can't actually solve for the number of moles of either gas without a temperature given so you'll just have to leave them in terms of T (I'll leave them in terms of RT actually).

Using n=PV/(RT) you have (1.7*2.7)/(RT) moles of H2 and (3.2*2.4/(RT) moles of He gas. These come out to ~4.6/(RT) moles of H2 and 7.7/(RT) moles of He for a total of 12.3/(RT) moles of gas.

So P = nRT/V leads us to P = (12.3/(RT))RT/(4.9L)

The RT cancels out and we're left with P = 12.3/4.9 ~ 2.5atm

Ultimately this works out the same if you took a sort of weighted average by using the following formula (P1V1+P2V2)/Vfinal.

(1.7*2.7+3.2*2.4)/4.9 ~ 2.5atm

The latter way is definitely faster but may not make as much intuitive sense.

2) The enthalpy change for the reaction below at 25oC is -1,449 kJ (per mole of C7H16). What is the internal energy change for the reaction at 25oC?
7 CO(g) + 15 H2(g) --> C7H16(l) + 7 H2O(l)

This last one is a thermodynamics question that is beyond the scope of the DAT in my opinion. The relationship between enthalpy and free energy is given by the following equation:

delta H = delta U + delta(PV) where delta U is change in internal energy

So delta U = delta H - delta(PV)

But according to the ideal gas law delta(PV) = delta(nRT)

and since R is a constant and T is fixed at a constant 25 C in this problem then

delta(nRT) = delta👎RT ultimately this just shows that only n is changing

So now delta U = delta H - delta👎RT

delta👎 is the change in moles of gas which is -22 in this case as we have 22 moles of gaseous reactants and zero moles of gaseous products.

delta U = -1449kJ - (-22)(8.314)(298)/1000 (dividing by 1000 converts to kJ)

delta U = -1449kJ - (-54.5kJ) = -1394.5kJ

But like I said, WAY BEYOND THE SCOPE OF THE DAT IMO!

 

[optomhopefully]

Thanks for clearing that up! Both of those answers make sense now! Since your an expert at these types of questions and this is clearly my weak point I was wondering if you didn't mind helping me with a few more I've tagged because they've been giving me trouble!

1. A 33.0-L cylinder contains 437 g O2(g) at 23.3°C. What mass of O2(g) must be released to reduce the pressure in the cylinder to 6.75 atm, assuming the temperature remains constant?

2. It takes 20.6 minutes for a 10.0-mL sample of an unknown gas to effuse through a pinhole. A 10.0-mL sample of helium, He, required 5.31 minutes. What is the molar mass of the unknown gas?

3. What is w when 1.79 kg of H2O(l), initially at 25.0oC, is converted into water vapour at 111oC against a constant external pressure of 1.00 atm? Assume that the vapour behaves ideally and that the density of liquid water is 1.00 g/mL.

4. A sample of gas absorbs exactly 443 J of heat from its surroundings. At the same time, the gas expands from 3.79 L to 12.71 L against a constant external pressure of 0.849 atm. What is the internal energy change for the gas?

5. The standard enthalpy change for the reaction below is -2391.7 kJ (for the reaction as written). What is the standard enthalpy of formation of WC(s)?
2 WC(s) + 5 O2(g) --> 2 WO3(g) + 2 CO2(g)

Data: Substance & Standard enthalpy of formation (in kJ mol-1)
WO3(s) -842.9
CO2(g) -393.5