I thought we went over this yesterday and you understood it. So what's the problem?
We didn't talk about the a1 part. For a1:
2O3 ==> 3O2 + heat.
Ke = (O2)^3/(O3)^2
increase in T.
So heat goes in on the right side, the reaction shifts to use the heat which means the reaction shifts to the left. If the reaction shifts to the left then concn O2 becomes smaller and concn O3 becomes larger. O2 is in the numerator, O3 is in the denominator. So if the upper number goes down and the lower number goes up, that means Ke MUST become either (smaller/larger)? Which one and you have the answer for a1. We discussed the other scenarios.
Did the equation do not know the 2nd questions"s answers for the problems
a) 2O3(g) = 3 O2(g) + heat (O3 IS UNDERLINED)
[6.0x10-1] [0.21]
b) 2 CO2 (g) + heat = 2 CO(g) +O2(g) (CO IS UNDERLINED)
[0.103] [0.024] [1.18 x10-2]
c) NO2(g) + O2(g) + heat =NO (g) + O3(g) (NO IS UNDERLINED)
[0.072] [0.083] [6.73X10-2] [6.73 X10-2]
.
For all three equilibria in problem #1 predict(1) how Ke is affected by an increase in temperature,(2)predict how the equilibrium will shift when pressure is decreased,(3)predict how the equilibrium will shift when the concentration of the underlined substance is increased ,and (4)predict how the equilibrium will shift when the temperature is decreased
1 answer