chem question help please

OP

Original post by Nitrotoluene

Calculating the Reaction Rate
Let’s find out how to figure out the rate of this reaction step by step. We’ll look at the hydrogen gas (H2) produced since it helps us understand how quickly the reaction is happening.
What we know:
- Hydrogen gas collected: 6.4 cm^3 in 60 seconds
- Molar volume of a gas (at SATP): 24.5 dm^3 per mole
a) : Convert the Volume
First, we need to change the hydrogen volume into dm^3 (because the molar volume is in dm^3).
Since 1000 cm^ equals 1 dm^3, we can do the conversion:
6.4 cm^3 = 6.4 / 1000 = 0.0064 dm^3
b) : Find moles of hydrogen
Now, we can use the molar volume to see how many moles of H2 were produced:
Moles of H2 = Volume / Molar Volume = 0.0064 dm^3 / 24.5 dm^3/mol = 2.61 × 10^-4 moles
c) : Reaction rate in mol/s
Next, let’s calculate how fast H2 is forming per second:
Rate = Moles of H2 / time = 2.61 × 10^-4 mol / 60 s = 4.35 × 10^-6 mol/s
For Experiments 1 and 2, repeat the same calculations.
My 2 cents!

Thank you, but the answer is given in mol dm-3 s-2 not mol/s, which is why I got confused. So don't we have to find concentration from the moles?

Reply 3

Original post by anonymous56754

Please could I have some help on this Question 2ci on page 7- I'm confused as to how they have worked it out. I decided to use 1mol=24000cm^-3 and then found the number of moles and then divided by volume to find concentration and then divided by 60. This is because the units is moldm-3s-1 but what the ms says seems to work it out as mols-1?
paper: https://www.physicsandmathstutor.com/pdf-pages/?pdf=https%3A%2F%2Fpmt.physicsandmathstutor.com%2Fdownload%2FChemistry%2FA-level%2FPast-Papers%2FCAIE%2FPaper-4%2FQP%2FMarch%202023%20(v2)%20QP.pdf
thanks!

Strictly speaking the units of molar volume are cm^3 mol^-1, but note that the column with the volume is in cm^3 / min

So “volume” divided by the molar volume will be in mol / min, hence the need to divide by 60 to convert the min to seconds.

As for where the dm^-3 comes from, there is no reason that can be inferred reliably in light of the information given. I mean if the solution had a volume of 1 dm^3, that would make sense as the number of moles of H2 made equals the number of moles of each reactant used up and therefore (ignoring the units), the changes in moles and changes in concentrations would be equal.

Reply 4

OP

Original post by TypicalNerd

Strictly speaking the units of molar volume are cm^3 mol^-1, but note that the column with the volume is in cm^3 / min
So “volume” divided by the molar volume will be in mol / min, hence the need to divide by 60 to convert the min to seconds.
As for where the dm^-3 comes from, there is no reason that can be inferred reliably in light of the information given. I mean if the solution had a volume of 1 dm^3, that would make sense as the number of moles of H2 made equals the number of moles of each reactant used up and therefore (ignoring the units), the changes in moles and changes in concentrations would be equal.

Are you saying that since it is a 1:1 ratio the change in moles is equal to change in concentrations so we can assume the solution has a volume of 1dm^3? Why can't we just calculate the concentration by dividing by volume 6.4*10-3dm-3? Thanks

Reply 5

Original post by anonymous56754

Are you saying that since it is a 1:1 ratio the change in moles is equal to change in concentrations so we can assume the solution has a volume of 1dm^3? Why can't we just calculate the concentration by dividing by volume 6.4*10-3dm-3? Thanks

I am saying that if x moles of H2 are made, x moles of each reactant are used up (because of the 1:1 ratio in the equation) and so IF and ONLY IF the volume of the solution is 1 dm^3, the concentrations of the reactants change by x mol dm^-3.

Rates of reaction strictly are a measure of the change in concentrations (usually of reactants, but sometimes products) over a period of time. This is why they are classically measured in mol dm^-3 s^-1.

Reply 6

OP

Original post by TypicalNerd

I am saying that if x moles of H2 are made, x moles of each reactant are used up (because of the 1:1 ratio in the equation) and so IF and ONLY IF the volume of the solution is 1 dm^3, the concentrations of the reactants change by x mol dm^-3.
Rates of reaction strictly are a measure of the change in concentrations (usually of reactants, but sometimes products) over a period of time. This is why they are classically measured in mol dm^-3 s^-1.

but the question doesn't tell us the volume is 1dm-3? Or thats just an assumption?

Reply 7

Original post by anonymous56754

but the question doesn't tell us the volume is 1dm-3? Or thats just an assumption?

Indeed it does not - it’s the only way I could see the units working out.

Reply 8

OP

Original post by TypicalNerd

Indeed it does not - it’s the only way I could see the units working out.

ok fair, but wouldn't dividing by the volume be more accurate?

(edited 3 weeks ago)

Reply 9