Chemistry MCQ thread...

A

Abdulrab

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just give me some idea as i mtotally confused in this question !!! u r genious please try it XD !!!
 
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sniper7137

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@Octahedral. Disproportionation means the same species being oxidized and reduced at the same time. The answer is A. Because:

1. Chlorine is being reduced to Cl- and oxidized to HClO at the same time.
2. Chlorine is being reduced to Cl- and oxidized to ClO3(-) at the same time.
3. NO2 is being reduced to NH3 and oxidized to HNO3 at the same time.

So 1,2,3 all apply to disproportionation. Hence the answer is A. Got it?
 
sweetiepie

sweetiepie

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fine if dint get it or m rong then m sorrieee jux read

hmm as shown in diagram s-nitrogen n i guess we have to tel gasses in bulb r cause sumthingg
so when temperature increase endothermic reaction favoured so itx positive n only one is posibble so da answer is D

I hope itx clear to u..:)
 
Mobeen

Mobeen

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Abdulrab said:
plzzzz !!! warna i ll be i tension !!! :( please give it a try !!! i will be thankful 2 u !!
Click to expand...
listen the question says we are increasing the temp.. so in reaction 1 :
the forward reaction is endothermic which means when we would incease the temp equilibrium would go to the right side.. the moles of product are double which means the vol would also double .. when more gas in the left side the mercury would increase in the right tube ...
in both reactions 2 and 3 the moles are the same for both sides thus the volumes are also same .. so the equilbrium wont cause any change in vol .. so ans is D ..
now did u get it .. i explained as much i could ... :D
 
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shyqueen

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What has tha same value as the standard enthalphy change of formation of carbon monoxide?
A. 1/2 enthalphy change of combustion (graphite)
B.enthalphy change of combustion(graphite)-enthalphy change of combustion(CO)
C.enthalphy change of formation (CO2)-enthalphy change of combustion(graphite
D.enthalphy change of formation (CO2)- 1/2 enthalphy change of combustion (graphite)
 
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sniper7137

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@Abdulrab. Here's what i think:

No matter what the number of moles in R and S, they exert the same pressure.

Now with increase in temperature:

1. The reaction is endothermic in fwd dir. So increase in temp. shifts the equil to the right. Means the reaction mixture now contains a greater number of the gas molecules. As the mole ratio is 1 : 2. More gas molecules means more pressure on the mercury causing it to rise to the right. So this is it.
2. The reaction is endothermic in fwd dir. But it doesn't matter cos no matter what the equil position the number of moles in R remain constant. So this is not it.
3. Nitrogen, when heated will exert more pressure but since the number of moles will be constant and temperature rise will be for both R and S the mercury level is unaffected.

Therefore only option 1 will lead to a rise in pressure on the left side leading to a rise in the height of the mercury on the right side. So thats why the ANSWER IS D. Got it?
 
Mobeen

Mobeen

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no problem .. now i should really go sleep .. havnt slept in two days .. then was out in the rain all the time . tired ..
but really this loooong thread has been really helpfull... thank you all guys .. bye ! :)
 
Saturation

Saturation

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sniper7137 said:
@Octahedral. Disproportionation means the same species being oxidized and reduced at the same time. The answer is A. Because:

1. Chlorine is being reduced to Cl- and oxidized to HClO at the same time.
2. Chlorine is being reduced to Cl- and oxidized to ClO3(-) at the same time.
3. NO2 is being reduced to NH3 and oxidized to HNO3 at the same time.

So 1,2,3 all apply to disproportionation. Hence the answer is A. Got it?
Click to expand...

For number 3, ammonia isn't being formed! It's HNO2 (nitrous acid) , which is still dispropotionation!
 
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