Cssa 04 (1 Viewer)

[hscwav2012]

Two Multiple Choice Questions need help with please

8. The pH of a 0.0115 mol/L solution of H₂SO₄ is closest to:
A - 0.9
B - 1.2
C - 1.6
D - 1.9

ANSWER-C

15. The Haber process is given below:
N_2(g) + 3H_2(g)

2NH_3(g)
Which of the following will favour the production of Ammonia?

A - Decreasing concentration of N_2(g)
B - Decreasing the pressure of the system
C - Adding Ar_(g)
D - Decreasing the volume of the system

ANSWER -D

 

[louielouiee]

For 15,
It's all about Le Chatelier's Principle.
By DECREASING the volume of a system in equilibrium, this pushes the equilibrium in the FORWARDS direction, favouring the FORMATION of the products, in this case Ammonia.

 

[Aesytic]

8. H2SO4 is a diprotic acid, so the concentration of hydrogen ions is twice the concentration of H2SO4
.'. [H+] = 2*[H2SO4]
= 2*0.0115
=0.023 mol/L
then using pH = -log[H+],
pH = 1.638 which is closest to C

15. with this equilibrium, we can see that there are 4 moles of gas on the LHS and 2 moles on the RHS
it doesn't matter what gas they are, since it is only the number of moles that affect the volume taken up
the answer is D, because if we decrease the total volume of the system, we are increasing the overall pressure in the system. think of it like having a plunger and pushing down - you decrease the volume, but pressure increases.
by Le Chatelier's Principle, the system will attempt to counterract this increase in pressure, and the way this is done is to shift the equilibrium to the side with less moles of gas to reduce the overall pressure on the system. the side with less moles in this case is the RHS, so we increase the yield of ammonia, making D the correct answer

 

[golgo13]

Just to bridge onto the answer for 15, when you decrease volume in the closed system for "GAS", it increases pressure, and by increasing pressure, you increase the collision rate of molecules thus favor the products

 

[someth1ng]

Just to bridge onto the answer for 15, when you decrease volume in the closed system for "GAS", it increases pressure, and by increasing pressure, you increase the collision rate of molecules thus favor the products

To be more precise, it favours the forward reaction as it produces less moles of gas reducing pressure, partially opposing the imposed changed of increase of pressure as according to Le Chatelier's principle.

The faster rate of collisions increases the speed of the reaction, but does not change the actual equilibrium (I think).

 

[golgo13]

You would be correct but i think i should of put more context into it. In a equilibrium system, say you suddenly change the volume (reduce it), then the presure suddenly increases, NOTE this is the part where the answer you supplied can be challenged, if the volume is reduce uniformly on both sides then there shouldn't be a change cause there is uniformity means that there isn't a real change. But however if the side of the reactants had an increase in pressure then yes the answer you have supplied will not be challenged.
I said it increased the speed of the reaction because i related it to industrial method, where they run high temperature (increase kinetic energy), and relatively high pressure. When you have less room the chances of finding say a nitrogen molecule with hydrogen molecules increases, thus wouldn't that be true the rate would increase?