2Pb(NO3)2 (s) = 2PbO (s) + 4NOs (g) + O2 (g) - I dont understand the effect on the position of the equilibrium
I dont understand why and which way they move when
NO2 is removed?
Solid PbO is added?
and
The pressure is increased?
Thanks guys,
Steve.
Answer:
2Pb(NO3)2 (s) = 2PbO (s) + 4NO2 (g) + O2 (g)
1.) NO2 is removed. Shift Forward. If you remove some NO2, what the system tends to do is to produce more NO2 to replace the lost NO2. Thus, it will favor a shift to the right.
2.) Solid PbO is added: It won't affect the equilibrium, because solids don't affect the equilibrium(ergo, theyre not in the Kc). Thus, No shift.
3.) Pressure is increased. If we increase the pressure, we decrease the volume of the container, thus, it tends to move to the side where there is "LESS" moles of gas, because only the volume of the gas is the volume of the system, thus, if you COMPRESS the area, the system will move to the area to relieve the stress. Thus a backward shift.
i think its 4NO2 not 4NOs...
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NO2 is removed?
Solid PbO is added?
and
The pressure is increased?
Thanks guys,
Steve.
Answer:
2Pb(NO3)2 (s) = 2PbO (s) + 4NO2 (g) + O2 (g)
1.) NO2 is removed. Shift Forward. If you remove some NO2, what the system tends to do is to produce more NO2 to replace the lost NO2. Thus, it will favor a shift to the right.
2.) Solid PbO is added: It won't affect the equilibrium, because solids don't affect the equilibrium(ergo, theyre not in the Kc). Thus, No shift.
3.) Pressure is increased. If we increase the pressure, we decrease the volume of the container, thus, it tends to move to the side where there is "LESS" moles of gas, because only the volume of the gas is the volume of the system, thus, if you COMPRESS the area, the system will move to the area to relieve the stress. Thus a backward shift.
i think its 4NO2 not 4NOs...
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