Question

If Keq = 4 x 10^-7, which of the following is true?
A) products are favored
B) reactants are favored
C) there are no reactants remaining
D) reactants and products are present in equal amounts


2. What does a chemical reaction rate measure?
A) the amount of reactant used
B) energy change per unit of time
C) change in amount of reactant per unit of time
D) the amount of time needed for the reaction to come to completion


3. In an equilibrium reaction that makes more moles of gas than it consumes, what is the effect of increasing the pressure?
A) the reaction makes more reactants
B) the reaction makes more products
C) the reaction is unchanged
D) the answer cannot be determined


4. What is the effect of adding ore water to the following equlibrium reaction? CO2 + H20 <--> H2CO3
A) more H2CO3 is produced
B) there is no effect
C) the equilibrium is pushed in the direction of the reactants
D) CO2 concentration increases


5. Which of the following explains the increase in reaction rate from a catalyst?
A) more collisions between particles, so more successful collisions
B) catalysts do not increase the reaction rate
C) same number of collisions, but more are successful
D) fewer collisions but more successful


6. Which of the changes listed below would shift the following reaction to the right? 4HCI(g) + 02(g) <--> 2CL2(g) + 2h20(g)
A) increase of pressure
B) addition of Cl2
C) decrease of pressure
D) removal of 02

Answer 1

1. Keq = product/reactant
If Keq is higher than 1, products are favored. And if lower, reactants are favored.

2. reaction rate measures the rate of the disappearance of a reactant or of the formation of a product.

3. increasing pressure increases the rate of the reaction

4. adding more water won't do anything because the reaction also requires more CO2 to produce more H2CO3

5. a catalyst increases the rate of reaction by lowering the activation energy. they do this by attracting the molecules of the reactants to its surface so that they are closer together and can react more easily by colliding more successfully.

6. increasing pressure would increase the rate but that does not matter since it's an equilibrium. adding chlorine wouldn't do anything too because you would need more water too. decreasing pressure, same thing as first one. removing O2 will shift the reaction to the right because there are less reactants on the left side of the reaction now.

Answer 2

Thank thomaster.

Answer 3

disagree on #6 (though i could be wrong). removing O2 will shift the reaction to the *left* precisely because there are less reactants.

to explain this. we could invoke lechatlier or just look at the reaction quotient, which has the oxygen in the denominator. if we decrease the partial pressure of O2, the reaction quotient will rise above the equilibrium constant, and the reaction will shift to the reactants side, so that the reaction quotient can become equal to the equilibrium constant again.

none of the choices sound right. if i have to pick "increase the pressure" with an important catch, we have to increase everyone's partial pressure by pressing on the reaction vessel. the reaction will then shift to the right, because for every mol of reaction, we convert 5 mol of reactant particles into 4 mol of product particles.

Answer 4

LOL @gsoda Soda increase in pressure is right please have the no of oles of gas on left and right :)

Answer 5

eh-hem...let me repeat what i said...if i have to pick a choice, i would pick "increase the pressure" with an important catch--we have to compress all of the gases. increasing the total pressure by injecting an inert gas won't shift the reaction to the right. See discussion on lechatlier in "Chemistry: the Central Science"

Answer 6

yaaa dude I doo ageree With you :) @gsoda wisky Scothch:)

Answer 7

Disagree with #4. It should be C). According to le Chatelier principle, it'll change the equilibrium equation, and push the reaction towards the right. The fact that it's a limiting reagent doesn't change the equilibrium equation.

Answer 8

For #6, removal of O2 would push the reaction to the left, to produce more O2, and bring back equilibrium. I don't think there's any right answer for that question. Addition of Cl2 would also push the reaction to the left...

Answer 9

more H2CO3 is produced it should be the answer

Answer 10

Err yes, you're right. I got mixed up. A) is the answer to 4#.

Answer 11

so where would you get the extra needed CO2 then? you only add more water. I don't see how you can produce more H2CO3 with only adding more water.

Answer 12

as far as 6 is concerned mee to have doubts but i think its increase of pressure :)LOL

Answer 13

i am not asking more CO2 but saying that it will move in right direction with increased rate compared to first one since it will be equllibrium always CO2 will be thee LOL in a

Answer 14

@thomaster It's an equilibrium reaction, so the final concentrations all revolve around the equilibrium constant and equation: [H2CO3]/[CO2][H20] = Ka (constant). If you add some water, the system has to compensate by creating more H2CO3 to arrive to the same constant. If you add more H2CO3, it'll create more CO2, vice-versa.

If it was a complete reaction (non-equilibrium), if CO2 is limiting, adding more water will do nothing.

Answer 15

@chmvijay Increase in pressure doesn't affect equilibrium. Only change in volume does, if there's a difference in the number of moles between reactants/products.

Answer 16

effect of pressure will be there when there's a difference in the number of moles between reactants/products.which we can see here