Chemistry: Post your doubts here!

[My Name]

Answer is A
Why? Why not C since the forward reaction is exothermic increasing temperature would the favour the reactants side right?

 

[My Name]

Answer is D
How?

 

[Xylferion]

X is a mixture of two compounds of Group II elements. X can undergo thermal decomposition to produce a white solid and only two gaseous products. One of the gaseous products relights a glowing splint. What could be the components of mixture X?
A MgCl 2 and CaCO3
B MgCO3 and Ca(NO3)2
C Mg(NO3)2 and Ca(NO3)2
D MgO and CaSO4
the answer is C... how can you distinguish between all the options and conclude that C is the right one? i suck at periodicity... any advice?

Group two elements undergo thermal decomposition like this:

2 X(NO3)2 ------> 2XO + 4NO2 + O2

X being any group 2 element.

The two gaseous products formed are nitrogen dioxide and oxygen, with XO being the white solid. Both Mg and Ca's oxides give white solids. The test for oxygen is that it will relight a glowing splint, so that too confirms that one of the products should be oxygen.

Therefore C is your answer.

 

[My Name]

How would I solve this? ,_,

 

[My Name]

Answer is C

 

[Xylferion]

View attachment 52140
Answer is A.
Explanation needed !
Please and Thank you.

As the reaction proceeds, the reactions will get used up and reactants will be formed. At the end of the reaction, there'll be no more reactants to form products from, so the graph has to flatten out at the end to indicate that no more [CH3OH] is going to be formed.

If you look at graph B, it's unusual for a reaction to proceed slowly and then all of a sudden just speed up and then slow down.

Graph C is also uncharacteristic since if no more [CH3OH] is being formed, how will the reaction carry out for more seconds without anything being produced.

In graph D, the reaction is not reversible, so where could all the CH3OH go unless we were to remove it? They asked for the variation with time of the amount of CH3OH formed as the reaction proceeds, we aren't supposed to intervene and alter the amounts produced.

Generally you want to have the graph flatten out as the reaction proceeds, keeping in mind that there will be a point where no more reactants are being made.

Hope that helped!

 

[Xylferion]

View attachment 52141
Answer is B

Forward reaction is endothermic. Endothermic reactions favour an increase in temperature, so since the forward reaction is endothermic, an increase in temperature shifts the equilibrium to the right.

With an increase in temperature, and since we're dealing with gases, the gas particles gain more kinetic energy and are able to spread out even further, which feeds into the whole concept of thermal expansion. The backwards reaction cannot counteract this because an increase in temperature has no business with the reverse reaction, which is exothermic.

Your only option then can be B, since as the gas particles spread out, the volume occupied by the gas increases.

 

[Xylferion]

View attachment 52142
Answer is A
Why? Why not C since the forward reaction is exothermic increasing temperature would the favour the reactants side right?

Increasing the temperature will increase the rates of both the forward and backward reactions, it's just that the backward reaction will increase at a faster rate.

Remember, the reaction is in equilibrium.

Equilibrium has to be dynamic, in that, both the forward and backward reactions must be equal. If you only have one increasing, there will be an imbalance and so no equilibrium would exist.

 

[My Name]

Increasing the temperature will increase the rates of both the forward and backward reactions, it's just that the backward reaction will increase at a faster rate.

Remember, the reaction is in equilibrium.

Equilibrium has to be dynamic, in that, both the forward and backward reactions must be equal. If you only have one increasing, there will be an imbalance and so no equilibrium would exist.

Equilibrium means that the rate of forward reaction is equal to the rate of backward reaction.
So they increased the temperature and still want it to be in Equilibrium?
When the temperature is increased the temperature more of products will be formed since forward reaction is exothermic....but since they want equilibrium both sides increase...
Did I get it right ?

 

[Xylferion]

View attachment 52143
Answer is D
How?

You need to consider the following:
Initially ---> Change -----> Equilibrium

Initially there are 2 moles of P and no products yet. As the reaction proceeds, P partly decomposes to produce Q and R.
At equilibrium they tell you that x moles of R were present and the total number of moles of both the reactants and products = (2 +x/2 )

You need to consider each of the 4 cases in order.

A) P <-> 2Q + R
I: 2 ---> 0 : 0
C: 2 - x ----> +2x : +x
E: 2 -x ---> 2x + x
Total moles: 2-x + 2x + x = 2 + 2x

-What this method basically does is consider what we have and what we end up with. You start with 2 moles, and at equilibrium x moles of R forms. So if you form products, you lose reactants. Which is why I did 2 - x, because I'm using up the reactants to form products.
- A cannot be the answer since what we're after is (2 + x/2), so repeat this for the other 3.

B) 2P <-> 2Q + R
I: 2 ---> 0 : 0
C: 2-2x ----> +2x : +x
E: 2-2x ---> 2x + x
Total moles = 2-2x + 2x + x = 2 + x

- This time since we have 2 moles of P at equilibrium, we will lose twice the amount. Hence the, 2-2x.

C) 2P <-> Q + R
I: 2 ---> 0 : 0
C: 2 - 2x ---> +x : +x
E: 2-2x ---> x + x
Total moles = 2

D) 2P <-> Q + 2R
I: 2 ---> 0 : 0
C: 2 -2x ---> +x : +2x

This whole time we were working off of the ratios of P:Q:R, keeping in mind that we need to form x moles of R.
Over here, we're forming 2x moles of R, we need to divide all of the changes by 2.

We end up with,
C: 2-x ----> +0.5x : +x
E: 2-x ---> 0.5x + x
Total moles = 2 + 0.5x = (2 + x/2 )

So your answer is D.

Hope that made sense!

 

[Xylferion]

View attachment 52145
How would I solve this? ,_,

For the reaction to proceed rapidly, the activation energy must be low. Since a lower activation energy means the particles gain sufficient energy faster.

The only graphs with low activation energies are C and D, but C has a lower activation energy than D, so your answer will be C.



Red lines are the activation energies.

 

[Xylferion]

Equilibrium means that the rate of forward reaction is equal to the rate of backward reaction.
So they increased the temperature and still want it to be in Equilibrium?
When the temperature is increased the temperature more of products will be formed since forward reaction is exothermic....but since they want equilibrium both sides increase...
Did I get it right ?

Both will increase, just one increases faster than the other to produce the yield faster.

 

[My Name]

Both will increase, just one increases faster than the other to produce the yield faster.

Why are BOTH increasing?

Thank you for your help!

 

[mohammed awadh]

Can someone please answer this question for ME PLEASEEEEEEEEEEEEEEE AND EXPLAINN THANKYOU



ANSWER IS B BUT YYYYYYYYYYYYYe

 

[Xylferion]

View attachment 52146
Answer is C

As the plunger is pushed in, the volume decreases and the pressure increases. Higher pressure favours a fewer number of moles. At equilibrium there's 1 mol PCl5 on the left compared to 2 moles of gases on the right. So as the plunger is pushed in, the position of the equilibrium shifts to the left, producing more PCl5.

If no dissociation were to happen, the reaction would not be in equilibrium. Hence only 2 moles of gas would be present at the end. Since it partly dissociated though, you can take into account the moles of PCl5, since that is present too. So you'd have 3 moles compared to 2 without dissociation. Which agrees with statement 3.

As for statement 1, I believe it's the pressure of the compressed gas that's greater but not what's in the syringe oven. Since the pressure of compressed air is greater than atmospheric pressure, I can't think of any other reason why the statement might be wrong. It has to be the wording.

Hope that helped.

 

[Xylferion]

Why are BOTH increasing?

Thank you for your help!

You already said why, because you still want the reaction to be in equilibrium.

 

[My Name]

You already said why, because you still want the reaction to be in equilibrium.

Yes...okay
So what I said here

Equilibrium means that the rate of forward reaction is equal to the rate of backward reaction.
So they increased the temperature and still want it to be in Equilibrium?
When the temperature is increased the temperature more of products will be formed since forward reaction is exothermic....but since they want equilibrium both sides increase...
Did I get it right ?

was right
Finally...I get it.... Thank you!

 

[Xylferion]

View attachment 52149


Can someone please answer this question for ME PLEASEEEEEEEEEEEEEEE AND EXPLAINN THANKYOU



ANSWER IS B BUT YYYYYYYYYYYYYe

You write out the bromination of propane.
Br2 ---> 2Br・

Br・ + C3H8 ---> C3H7・ + HBr

C3H7・ + Br2 ----> C3H7Br + Br・

Br・ + C3H8 ----> C3H7 ・ + HBr

Then in the termination step, you have two radicals react. So you can have C3H7・ and C3H7・ react to form C6H14.

The catch here is that there's two types of radicals that react. It may look like it's just C3H7・, but the ・ can be attached to 2 different carbons in C3H7. Either the first or the second.

Like so,

CH3CH2CH2・

Or

CH3CH・CH3

Option 1 follows,

CH3CH2CH2・ + CH3CH・CH3

While option 2 follows,

CH3CH・CH3 + CH3CH・CH3

Option 3 cannot be formed because no combination of CH3CH・CH3 and CH3CH2CH2・ would allow for it.

Hope that made sense!

 

[mohammed awadh]

You write out the bromination of propane.
Br2 ---> 2Br・

Br・ + C3H8 ---> C3H7・ + HBr

C3H7・ + Br2 ----> C3H7Br + Br・

Br・ + C3H8 ----> C3H7 ・ + HBr

Then in the termination step, you have two radicals react. So you can have C3H7・ and C3H7・ react to form C6H14.

The catch here is that there's two types of radicals that react. It may look like it's just C3H7・, but the ・ can be attached to 2 different carbons in C3H7. Either the first or the second.

Like so,

CH3CH2CH2・

Or

CH3CH・CH3

Option 1 follows,

CH3CH2CH2・ + CH3CH・CH3

While option 2 follows,

CH3CH・CH3 + CH3CH・CH3

Option 3 cannot be formed because no combination of CH3CH・CH3 and CH3CH2CH2・ would allow for it.

Hope that made sense!

im so weak in this . do u have an easy explaination ? i understood the radical part but applying it was difficult.
thanks really for the tip

 

[Xylferion]

im so weak in this . do u have an easy explaination ? i understood the radical part but applying it was difficult.
thanks really for the tip

With propane, there are 3 carbons. You can have the little dot ・, which is basically an electron, appear on any of the three carbons.

The thing is, if you have it on the first and third carbon, that's the same thing, because you will start numbering the carbons, from the one that has ・.

The only other carbon which you can have the ・ attached to, is the second one.

This gives you only two places to put ・.

During the termination step, you have two combine two radicals. That is two molecules that have ・.

Since ・ can be on the first OR the second carbon, you can form 3 combinations.

First carbon radical ・ + First Carbon radical ・
First carbon radical ・ + Second carbon radical ・
Second carbon radical ・ + Second carbon radical ・

So if you have C3H7・ as your radical. The ・ can be on the first or second carbons.

-C-C・-C-
OR
・C-C-C-