chem mcq help please

Hi, please could I have help on these two questions:

1. I know how to do the calculation but I am pretty sure the answer is wrong. Since volume is in m3, the pressure will be in pascals but to convert to kilopascals, dividing by 1000 gives me 1.08*10-5 which isn't an option?
2. I don't understand the difference between 1 and 2, aren't they both correct?
Here is the question: https://ibb.co/0jqBC5C
Thank you!

Reply 1

Nitrotoluene

Original post by anonymous56754

Hi, please could I have help on these two questions:
1. I know how to do the calculation but I am pretty sure the answer is wrong. Since volume is in m3, the pressure will be in pascals but to convert to kilopascals, dividing by 1000 gives me 1.08*10-5 which isn't an option?
2. I don't understand the difference between 1 and 2, aren't they both correct?
Here is the question: https://ibb.co/0jqBC5C
Thank you!

You are right! IMHO...of course!😀

Molar Mass of O2: 32 g/mol
Mass of O2 : 0.96 g
Temperature in °C: 30 °C
Volume V: 7.0 × 10^3 m^3
Ideal Gas Constant (R): 8.314 J/(mol x K)
Calculate Moles of O2
n = mass of O2/molar mass of O2 = 0.96 g/32 g/mol = 0.03 mol
Convert Temperature to Kelvin
To convert temperature from Celsius to Kelvin:

T = 30 °C + 273 = 303 K

Calculate Pressure using Ideal Gas Law

P = nRT/V

Solving for P and substituting the values:

P = (0.03 mol) × (8.314 J/(mol x K) × (303 K)/(7.0 × 10^3 m^3)

Calculating the numerator:

(0.03 × 8.314 × 303) = 0.03 × 2510.842 = 75.32526 J

Calculating the pressure (P):

P = 75.32526 J/7000 m^3 = 0.01076 J/m^3

Convert Pressure to kPa

Since 1 J/m³ = 1 Pa and 1 kPa = 1000 Pa: ==> P = 0.01076 Pa/1000 = 0.00001076 kPa = apx 1.08 x 10^-5 kPa

The calculated pressure (P) is approximately 0.00001076 kPa = apx 1.08 x 10^-5 kPa

The reason why the pressure is so low is that the volume is very large (7000 litres) and the mass of the O2 is very small (0.96 g). With a volume of 7 m^3, or 7,000 litres, imagine how high and how large a tank containing 0.96 grams of oxygen could be.

(edited 1 month ago)

Reply 2

TypicalNerd

Original post by anonymous56754

There’s a misprint in the question. Notice how there’s a space between the 10 and the power of 3 in the volume? There should be a minus sign so it should be a volume of 7 x 10^-3 m^3

For the second question, you are comparing two experiments where the main thing that is different is the volume of the container. As such the concentrations are different - in fact all the concentrations decrease because the volume is larger and the numbers of moles of each species are presumably quite similar. So which answer must be right?

(edited 1 month ago)

Reply 3

Nitrotoluene

Let us do the calculations again, taking into account TypicalNerd's input about the space between the 10 and the power of 3 in the volume. If there is a minus sign, the volume is 7 x 10^-3 m^3.
Since the volume V is actually 7 x 10^-3 m^3, not 7 x 10^3 m^3, we need to recalculate the pressure.
First, let's recalculate the moles of O2:

n = mass of O2/molar mass of O2 = 0.96 g/32 g/mol = 0.03 mol

Next, let's convert the temperature to Kelvin:

T = 30 °C + 273 = 303 K

Now, let's calculate the pressure using the Ideal Gas Law:

P = nRT/V

Substituting the values:

P = (0.03 mol) × (8.314 J/(mol x K) × (303 K)/(7.0 × 10^-3 m^3)

Calculating the numerator:

(0.03 × 8.314 × 303) = 0.03 × 2510.842 = 75.32526 J

Calculating the pressure P:

P = 75.32526 J/(7.0 × 10^-3 m^3) = 10761.88 Pa

Converting the pressure to kPa:

P = 10761.88 Pa/1000 = 10.762 kPa

The calculated pressure P is therefore 10.762 kPa = apx 10.8 kPa, which confirms the result C given in the text of this problem.

Bye,
Sandro

(edited 1 month ago)

Reply 4

anonymous56754

Original post by Nitrotoluene

You are right! IMHO...of course!😀
Molar Mass of O2: 32 g/mol
Mass of O2 : 0.96 g
Temperature in °C: 30 °C
Volume V: 7.0 × 10^3 m^3
Ideal Gas Constant (R): 8.314 J/(mol x K)
Calculate Moles of O2
n = mass of O2/molar mass of O2 = 0.96 g/32 g/mol = 0.03 mol
Convert Temperature to Kelvin
To convert temperature from Celsius to Kelvin:

T = 30 °C + 273 = 303 K

Calculate Pressure using Ideal Gas Law

P = nRT/V

Solving for P and substituting the values:

P = (0.03 mol) × (8.314 J/(mol x K) × (303 K)/(7.0 × 10^3 m^3)

Calculating the numerator:

(0.03 × 8.314 × 303) = 0.03 × 2510.842 = 75.32526 J

Calculating the pressure (P):

P = 75.32526 J/7000 m^3 = 0.01076 J/m^3

Convert Pressure to kPa

Since 1 J/m³ = 1 Pa and 1 kPa = 1000 Pa: ==> P = 0.01076 Pa/1000 = 0.00001076 kPa = apx 1.08 x 10^-5 kPa

The calculated pressure (P) is approximately 0.00001076 kPa = apx 1.08 x 10^-5 kPa
The reason why the pressure is so low is that the volume is very large (7000 litres) and the mass of the O2 is very small (0.96 g). With a volume of 7 m^3, or 7,000 litres, imagine how high and how large a tank containing 0.96 grams of oxygen could be.

Yes, exactly my thoughts thank you for the detailed explanation :smile:

Reply 5

anonymous56754

Original post by TypicalNerd

There’s a misprint in the question. Notice how there’s a space between the 10 and the power of 3 in the volume? There should be a minus sign so it should be a volume of 7 x 10^-3 m^3
For the second question, you are comparing two experiments where the main thing that is different is the volume of the container. As such the concentrations are different - in fact all the concentrations decrease because the volume is larger and the numbers of moles of each species are presumably quite similar. So which answer must be right?

Ah thank you for question 1, I didn’t spot that! For question 2, isn’t the forward reaction being favoured so the moles of the products are going to increase so the concentration will increase?

Reply 6

TypicalNerd

Original post by anonymous56754

Ah thank you for question 1, I didn’t spot that! For question 2, isn’t the forward reaction being favoured so the moles of the products are going to increase so the concentration will increase?

I think it’s more likely the concentrations will all decrease, given how much larger the flask is.

I also don’t see anything indicating what side of the equilibrium everything lies on. All I see is the temperature (600 K in both reactions) and that they used 1 mol of PCl5 in either a 1 dm^3 flask or a 2 dm^3 flask.

Let’s say that at equilibrium there are x moles of PCl3 (and so there are also x moles of Cl2 and 1 - x moles of PCl5) in experiment 1 and that there are y moles of PCl3 in experiment 2. Thus the we can say:

In experiment 1:

[PCl3] = [Cl2] = x M and [PCl5] = (1 - x) M

In experiment 2:

[PCl3] = [Cl2] = y/2 M and [PCl5] = (0.5 - y/2) M

Let’s say for argument’s sake that the equilibrium constant Kc = 1 at 600 K and therefore

x^2 / (1 - x) = 1 => x = 0.618 (by forming and solving a quadratic, then rejecting the negative solution)

Thus [PCl3] = [Cl2] = 0.618 M and [PCl5] = 0.382 M in experiment 1

(y/2)^2 / (0.5 - y/2) = 1 => y = 0.732 (similar process to the above, but more workings are needed)

Thus [PCl3] = [Cl2] = 0.366 M and [PCl5] = 0.134 M in experiment 2

If this doesn’t convince you, you can retry the calculations again, picking different values of Kc (as long as they are greater than 0) and you’ll find that in experiment 2, all the concentrations decrease because the effect of doubling the volume of the flask roughly halves the concentrations.

Reply 7

Nitrotoluene

2^nd problem
The equilibrium concentrations of PCl3(g) and Cl2(g) are higher in the second experiment because doubling the volume while keeping the amount of PCl5 constant reduces the initial concentration, shifting the equilibrium to the side with more moles of gas (PCl3 and Cl2).
Answer: A

Krgds,
Sandro

Reply 8

TypicalNerd

Original post by Nitrotoluene

Disagreed - pick a value of Kc (1 is a sensible guess) and try calculating the concentrations for yourself. I have already done the calculations in an earlier post and shown that the equilibrium doesn’t shift far enough for that to be the case.

Furthermore, if statement A is true, then that would imply that B must also be true as the law of conservation of mass would require [PCl5] to decrease if [PCl3] and [Cl2] were to be raised. There cannot be more than one right answer…

Only B is a correct answer - the concentrations of every species must decrease.

(edited 1 month ago)

Reply 9

anonymous56754

Original post by TypicalNerd

I think it’s more likely the concentrations will all decrease, given how much larger the flask is.
I also don’t see anything indicating what side of the equilibrium everything lies on. All I see is the temperature (600 K in both reactions) and that they used 1 mol of PCl5 in either a 1 dm^3 flask or a 2 dm^3 flask.
Let’s say that at equilibrium there are x moles of PCl3 (and so there are also x moles of Cl2 and 1 - x moles of PCl5) in experiment 1 and that there are y moles of PCl3 in experiment 2. Thus the we can say:
In experiment 1:
[PCl3] = [Cl2] = x M and [PCl5] = (1 - x) M
In experiment 2:
[PCl3] = [Cl2] = y/2 M and [PCl5] = (0.5 - y/2) M
Let’s say for argument’s sake that the equilibrium constant Kc = 1 at 600 K and therefore
x^2 / (1 - x) = 1 => x = 0.618 (by forming and solving a quadratic, then rejecting the negative solution)
Thus [PCl3] = [Cl2] = 0.618 M and [PCl5] = 0.382 M in experiment 1
(y/2)^2 / (0.5 - y/2) = 1 => y = 0.732 (similar process to the above, but more workings are needed)
Thus [PCl3] = [Cl2] = 0.366 M and [PCl5] = 0.134 M in experiment 2
If this doesn’t convince you, you can retry the calculations again, picking different values of Kc (as long as they are greater than 0) and you’ll find that in experiment 2, all the concentrations decrease because the effect of doubling the volume of the flask roughly halves the concentrations.

Thank you so much for this! The reason why I thought the equilibrium shifts to the right and the concentration increases is because the volume of the flask is increasing so the pressure decreases and due to le chateliers, the forward reaction is favoured? I understand your calculations but how do we know when to use le chateliers and when the volume increasing causing the concentration to decrease outweighs the former? Also for the moles of pcl5 wouldn’t it be 1-2x? Thank you.

Reply 10

TypicalNerd

Original post by anonymous56754

I think in general if the volume of the flask is the only thing that is changed, it always outweighs Le Chatelier if and only if you are comparing concentrations. It is highly unlikely they’d get you to sus out the effect of multiple changes (though they could give you a much simpler equilibrium that does not require forming a quadratic and calculating for yourself the effect on the concentration).

It is 1 - x moles of PCl5. If you look at the molar ratio of PCl5 used up to the moles of either PCl3 or Cl2 made, the ratio is 1:1. As such, the moles of PCl5 will decrease by the same amount the moles of either PCl3 or Cl2 will increase by, so if x moles of either gas is formed (when there is initially none of either), 1 - x moles of PCl5 must be left.

Reply 11

anonymous56754

Original post by TypicalNerd

I think in general if the volume of the flask is the only thing that is changed, it always outweighs Le Chatelier if and only if you are comparing concentrations. It is highly unlikely they’d get you to sus out the effect of multiple changes (though they could give you a much simpler equilibrium that does not require forming a quadratic and calculating for yourself the effect on the concentration).
It is 1 - x moles of PCl5. If you look at the molar ratio of PCl5 used up to the moles of either PCl3 or Cl2 made, the ratio is 1:1. As such, the moles of PCl5 will decrease by the same amount the moles of either PCl3 or Cl2 will increase by, so if x moles of either gas is formed (when there is initially none of either), 1 - x moles of PCl5 must be left.