1. The following reaction occurs in a closed container: A(g)+2B(g)>>2C(g) A reac

Question

1.

The following reaction occurs in a closed container: A(g)+2B(g)>>2C(g) A reaction mixture initially contains 1.3L of A and 2.2L of B.

Assuming that the volume and temperature of the reaction mixture remain constant, what is the percent change in pressure if the reaction goes to completion?

2.

Air is about 78.0% nitrogen molecules and 21.0% oxygen molecules. Several other gases make up the remaining 1% of air molecules

An "empty" container is not really empty if it contains air. How may moles of nitrogen are in an "empty" two-liter cola bottle at atmospheric pressure and room temperature 25 celsius Assume ideal behavior.

3.

A 1-L flask is filled with 1.20g of argon at 25C. A sample of ethane vapor is added to the same flask until the total pressure is 1.10atm.

What is the partial pressure of argon in the flask?

What is the partial pressure of ethane in the flask?

4. A 3.00-L flask is filled with gaseous ammonia, NH 3 The gas pressure measured at 24.0C is 1.85atm

Assuming ideal gas behavior, how many grams of ammonia are in the flask

4. (part B)

If 1.00 mol of argon is placed in a 0.500-L container at 30.0C, what is the difference between the ideal pressure (as predicted by the ideal gas law) and the real pressure (as predicted by the van der Waals equation)?

For argon, a=1.345(L 2 %u22C5atm)/mol 2   and b=0.03219L/mol .

Explanation / Answer

1)Because of the mole ratio,

1.3L A will require 2.6L of B, so you run out of B before all the A is reacted.

Starting with
2.2L of B will require 1.1L of A and produce 2.2L of C

Before the reaction there is 1.3L A and 2.2L of B for a total of 3.5L. So V is 3.5L
After the reaction there is
1.3 - 1.1 = 0.2L of A
0L of B and
2.2L of C.
for a total volume of 2.4L

But since the V is constant it is the pressure that changes.
The pressure will decrease since there is less gas so
P2 = P1 x 2.4 / 3.5 = .686 P1

(P - 0.686P) / P = 0.314 = 31.4%
(the negative sign means that the pressure decreased.)

2)aPV = nRT

Rearrange to solve for moles (n)

PV/RT = n

For nitrogen:

(0.78 atm)(2 L)/(0.082 L*atm*K-1*mol-1)(298.15K) = n

0.064 mol N2 = n

3)Ptotal = Pargon + Pethane

PV = nRT

P = nRT/V

Pargon = (1.00g x 1mole/39.95g) x (0.08206 Latm/moleK) x (298.15K) / (1L) = 0.61atm

Pethane = 1.10atm - 0.61atm = 0.49atm

4)a

Using PV = nRT,

n = PV/RT = (1.85 * 3)/0.0821*297.15 = 0.2275 moles

Now, all you need is to turn moles into grams: Grams ammonia = Grams = 0.2275 moles * 17.03 g/mole = 3.87 grams.

4)b

The Van der Waals equation is as follows (I already plugged in n = 1 mole for simplicity sake...):

P = RT/(V-b) - a/V^2

Where:

P is the pressure (what we are looking for)

V is the volume (0.500L in this question)

R is the gas constant (0.082 [L%u2009atm] /%u2009[K%u2009mol])

T is the temperature in Kelvin (303.15K)

a is a Van der Waals constant (for argon it is 1.355 [L^2 bar]/mol^2)

b is a Van der Waals constant (for argon it is 0.0320 L/mol)

But recall that we already plugged into the equation n=1 mole hence the unit of mole cancel out in our Van der Waals constants and the gas constant and we get:

a = 1.355 L^2 bar = 1.355 L^2 bar * 0.986923267 atm/bar = 1.337 L^2 atm

b = 0.0320 L

R = 0.082 [L%u2009atm] /%u2009K

Plugging those into the equation we get:

P = (0.082 [(L%u2009atm)/K] /%u2009* 303.15K) / (0.500L - 0.0320L) - [1.337 L^2 atm] / (0.500L)^2

= [24.8583 (L%u2009atm)] / 0.468L - [1.337 L^2 atm] / 0.25 L^2 =

53.116 atm - 5.348 atm = 47.8 atm (3 sig figs)

So we got 49.7 atm from the ideal gas law and we got 47.8 atm from the Van der Waals equation.

The difference is:

49.7 atm - 47.8 atm = 1.9 atm

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