VCE Chemistry Question Thread

[huity]

Thanks sweetcheeks. Got a couple more questions here. Sorry for the series of questions. For Q26, how would I identify what the reactant is?
For Q29, are we even able to attempt the question. I don't think enough imformation has been provided.

Don't worry about the questions, we're here to help 
Q26 For calculation questions, unless otherwise stated, assume that coal is 100% Carbon. Note: As you know from Ch1 of fuels though, carbon often contains nitrogen, sulfur, etc. too!

Q29 We know the delta H for one mol of the fuel. This is the same energy (kJ) released for 3 mols of CO2. (Think of stoich mol ratios  ). Then, you can use ratio to find out the energy released for 1 tonne of CO2. Hope that helps! Give it a go and if you're still stuck, I'll write out the working 

[lelouchoftherebellion]

Hi all, I believe I'm right in thinking A is the answer for this question. I just can't quite confirm my logic behind it, can anyone help?

[sweetcheeks]

Hi all, I believe I'm right in thinking A is the answer for this question. I just can't quite confirm my logic behind it, can anyone help?

I believe that B should be the correct answer (although the answer really should be 2135 kJ/mol rather than just kJ). The activation energy for the forward reaction of CH4 + 2O2 --> CO2 + 2H2O is 3380 kJ/mol and the enthalpy of combustion is -890 kJ/mol (from data booklet).

What this means is that the activation energy of the reverse reaction CO2 + 2H2O --> CH4 + 2O2 is going to be the activation energy of the forward reaction (3380 kJ/mol) minus the change in enthalpy (-890 kJ/mol). This gives us 3380 - (-890) = 4270 kJ/mol as the activation energy for CO2 + 2H2O --> CH4 + 2O2.

Since the coefficients have been halved, we halve the value of activation energy, giving 2135 kJ/mol of activation energy for the reaction 1/2CO2 + H2O --> 1/2CH4 + O2.

[dream chaser]

Thanks sweetcheeks. Got a couple more questions here. Sorry for the series of questions. For Q26, how would I identify what the reactant is?
For Q29, are we even able to attempt the question. I don't think enough imformation has been provided.

Could someone please show me the workings for Q26 and Q29. I'm still not sure how to attempt them. Thanks

[Yertle the Turtle]

Could someone please show me the workings for Q26 and Q29. I'm still not sure how to attempt them. Thanks

I think that this working should be correct. If not I'm sorry and someone else feel free to correct me.

[dream chaser]

I think that this working should be correct. If not I'm sorry and someone else feel free to correct me.

Thanks Yertle the Turtle. Your help is much appreciated.

I'm still not sure what the difference between 'Volume of CO2 produced' and 'Volume of CO2 released' and why the calculations to calculate them will be different. Could someone please explain it. What it says in my book will be in the attachment. Also, why would energy and volume of CO2 be released when they are the products?

[Yertle the Turtle]

Thanks Yertle the Turtle. Your help is much appreciated.

I'm still not sure what the difference between 'Volume of CO2 produced' and 'Volume of CO2 released' and why the calculations to calculate them will be different. Could someone please explain it. What it says in my book will be in the attachment. Also, why would energy and volume of CO2 be released when they are the products?

You seem to have missed what the different calculations refer to. The first one is "Volume of CO2 produced per kilogram of reactant" while the second one is "Volume of CO2 released per MJ of energy obtained". Thus they are possibly just different words for the same thing, but the point is not those different words, but the different conditions you are measuring from.

[gwenstacy]

What is the difference between energy content, enthalpy change and heat of combustion?

I am currently planning for my AOS 3 practical and getting these terms mixed up!

Help would be gladly appreciated

[EllingtonFeint]

Hi, I need help with this question. I'm getting a different answer to my textbook...

0.25 mol of nitrogen is placed in a flask of volume 5.0L at a temperature of 5.0°C. What is the pressure in the flask?

So, I used the formula

P = nRT/V

--> p= 0.25 * 8.31 * 278 / 5

to get an answer of 115 kPA. But the textbook's answer is 1.2 * 10^2 kPa.

Where am I going wrong??

[sweetiepi]

Hi, I need help with this question. I'm getting a different answer to my textbook...

0.25 mol of nitrogen is placed in a flask of volume 5.0L at a temperature of 5.0°C. What is the pressure in the flask?

So, I used the formula

P = nRT/V

--> p= 0.25 * 8.31 * 278 / 5

to get an answer of 115 kPA. But the textbook's answer is 1.2 * 10^2 kPa.

Where am I going wrong??

I believe that you're fine!
It appears that the textbook has just has their answer to two significant figures (as that's the least amount of sigfigs in the question)- rounding up 115 into 120 and then converting it into scientific notation to get \(1.2 \times 10^2\)

Hope this helps!

[dream chaser]

Hi Guys,

Could someone please explain and show their workings as to how to do this question. The question is attached below this post. I'm clueless as to how to attempt it.

All help will be much appreciated. Thanks.

[Yertle the Turtle]

Hi Guys,

Could someone please explain and show their workings as to how to do this question. The question is attached below this post. I'm clueless as to how to attempt it.

All help will be much appreciated. Thanks.

Basically for this question you would take your delta H value and use it to find how many times the reaction needs to be done to generate 1MJ of energy, and then multiply by the number of moles of carbon dioxide in the equation to find the number of moles of CO2 generated. Then you use n=V/V_m to find the volume of CO2 generated. Hope this helps.

[huity]

Exactly what Yertle the Turtle said! I've written out the working too, hope it helps 

2 mols of C2H6 release 3120 kJ.
So 0.641 mols of C2H6 release 1MJ (1000 kJ).
Since n(C2H6): n(CO2) is 2:4, 0.641*2=1.282 mols of CO2 release 1MJ.
Rearranging n=V/Vm, we get V=n*Vm. So V(CO2)= n(CO2) * 24.8 = 31.8 L

[persistent_insomniac]

What is the difference between the q in the formula q = mass x heat of combustion and q = mass x specific heat capacity x change in temp? Would you get the same answer with both formulas. My understanding is that the q in q = mcdeltaT is the energy transferred to the water/absorbed by water and q in q = m x Hc is the energy released by the combustion of the fuel.

[Yertle the Turtle]

What is the difference between the q in the formula q = mass x heat of combustion and q = mass x specific heat capacity x change in temp? Would you get the same answer with both formulas. My understanding is that the q in q = mcdeltaT is the energy transferred to the water/absorbed by water and q in q = m x Hc is the energy released by the combustion of the fuel.

The q in the first formula is the energy released by the combustion of the fuel, while the q in the second formula is the energy absorbed by the water, as you have said. However, even though, in VCE Chemistry, you would say that these two things are the same, they are actually not. Some amount of energy is released into the atmosphere, and thus the two q terms are not exactly the same. However, for the questions you will face in VCE Chemistry you are meant to take them as equivalent.