Chemistry: Post your doubts here!

[a_wiserME!!]

Already explained this. It's the first post on this page.

k thnk u

 

[a_wiserME!!]

can someone pls answer this???

 

[Xylferion]

View attachment 52135

can someone pls answer this???

The graph represents a two step reaction because of the two dips visible. So from all of those 3 options you need to eliminate the single step reactions.

All of the options presented are of halogenoalkanes under going substitution reactions with NaOH or NH3.

A key factor is the degree of each, that is whether they are primary, secondary or tertiary.

Option 1, the Carbon holding the Halide ion is attached to 3 other carbons, this makes it a tertiary halogenoalkane. These undergo two-step SN1 reactions that are split into a slow stage and then a fast stage.

Option 2, the Carbon holding the Halide ion is attached to 1 other carbon, making it a primary halogenoalkane. These undergo single-step SN2 reactions in which there is a single transition state.

Option 3 is another example of a primary halogenoalkane that will undergo a SN2 reaction.

So from all of the above cases, Option 1 is the only reaction that proceeds under two different stages, a slow and a fast one, which are distinguishable by the steepness of the gradients on the graph.

So your answer will be D.

 

[Xaptor16]

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?

 

[a_wiserME!!]

The graph represents a two step reaction because of the two dips visible. So from all of those 3 options you need to eliminate the single step reactions.

All of the options presented are of halogenoalkanes under going substitution reactions with NaOH or NH3.

A key factor is the degree of each, that is whether they are primary, secondary or tertiary.

Option 1, the Carbon holding the Halide ion is attached to 3 other carbons, this makes it a tertiary halogenoalkane. These undergo two-step SN1 reactions that are split into a slow stage and then a fast stage.

Option 2, the Carbon holding the Halide ion is attached to 1 other carbon, making it a primary halogenoalkane. These undergo single-step SN2 reactions in which there is a single transition state.

Option 3 is another example of a primary halogenoalkane that will undergo a SN2 reaction.

So from all of the above cases, Option 1 is the only reaction that proceeds under two different stages, a slow and a fast one, which are distinguishable by the steepness of the gradients on the graph.

So your answer will be D.

thnx once again

 

[My Name]

Answer is B

 

[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