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]
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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
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.
To answer these questions, we need to consider Le Chatelier's principle, which states that when a system in equilibrium is subjected to a change in temperature, pressure, or concentration, the system will shift in a way that counteracts the change and reestablishes equilibrium.
Let's go through each question one by one:
a) 2O3(g) = 3 O2(g) + heat (O3 is underlined)
1) When the temperature is increased, the equilibrium will shift in the direction that absorbs heat. In this case, the forward reaction is exothermic, meaning it releases heat. Therefore, increasing the temperature will shift the equilibrium to the left, favoring the reactants. Ke will decrease.
2) When the pressure is decreased, it will not affect this equilibrium since there are no gaseous reactants or products.
3) Increasing the concentration of O3 will shift the equilibrium to the left, favoring the reactants. It will cause the reaction to produce more O3 to reestablish equilibrium. Ke will decrease.
4) Decreasing the temperature will shift the equilibrium to the right, favoring the products. In an exothermic reaction, decreasing the temperature will cause the reaction to produce more heat to reestablish equilibrium. Ke will increase.
b) 2 CO2 (g) + heat = 2 CO(g) + O2(g) (CO is underlined)
1) Increasing the temperature will shift the equilibrium to the right, favoring the products. In an endothermic reaction, increasing the temperature will cause the reaction to absorb more heat to reestablish equilibrium. Ke will increase.
2) When the pressure is decreased, it will not affect this equilibrium since there are no gaseous reactants or products.
3) Increasing the concentration of CO will shift the equilibrium to the left, favoring the reactants. It will cause the reaction to produce more CO2 to reestablish equilibrium. Ke will decrease.
4) Decreasing the temperature will shift the equilibrium to the left, favoring the reactants. In an endothermic reaction, decreasing the temperature will cause the reaction to release more heat to reestablish equilibrium. Ke will decrease.
c) NO2(g) + O2(g) + heat = NO(g) + O3(g) (NO is underlined)
1) Increasing the temperature will shift the equilibrium to the right, favoring the products. In an endothermic reaction, increasing the temperature will cause the reaction to absorb more heat to reestablish equilibrium. Ke will increase.
2) When the pressure is decreased, it will not affect this equilibrium since there are no gaseous reactants or products.
3) Increasing the concentration of NO will shift the equilibrium to the left, favoring the reactants. It will cause the reaction to produce more NO2 and O2 to reestablish equilibrium. Ke will decrease.
4) Decreasing the temperature will shift the equilibrium to the left, favoring the reactants. In an endothermic reaction, decreasing the temperature will cause the reaction to release more heat to reestablish equilibrium. Ke will decrease.