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CO(g) + H2O(g) <--->CO2((g) + H2(g) 

kc= 7.3

delta H= -41.2KJ

if the temperature of the system at equilibrium is increased, then the concentration of the H2(g) will _______ and the value of kc will_____

for the blanks choose one from the following 

- increase

-decrease

-stay the same 

 

question 2 

Solid copper(II) sulfate was added to a flask containing water and then stoppered. the following solubility equilibrium was established . 

CuSO4(s) <---> Cu2+(aq)+ SO42-(aq)

which of the following observations would indicate that equilibrium has been established?

1. the forward and the reverse reaction have stopped

2. the mass of the solid copper(II) sulfate is completely dissovled

3. the pressure of the system is constant

4. the mass of solid copper(II) sulfate remaining in the flask is constant

 

5 years ago

Answered By Leonardo F

Question 1) The first concept that we need to remember is that if the deltaH of the forward reaction is negative, this means that the forward reaction is exothermic (releases energy) and, so, the reverse reaction is endothermic (absorbs energy). In a established chemical equilibrium, a temperature rise in the system shifts the equilibrium to the endothermic side, while a temperature drop shifts it to the exothermic side. Since the exercise told us that the temperature is increased, that means that we shift the equilibrium to the left (reverse reaction - endothermic). This means that the concentration of  $CO_2$CO2 and  $H_2$H2 will decrease, while the concentration of  $CO$CO and  $H_2O$H2O will increase.

So, the first blank is "decrease". Let's now express kc:

 $kc=\frac{\left[CO_2\right]\left[H_2\right]}{\left[CO\right]\left[H_2O\right]}$kc=[CO2][H2][CO][H2O]  

Since the concentration of $CO_2$CO2 and  $H_2$H2 will decrease, while the concentration of  $CO$CO and  $H_2O$H2O will increase, the value of kc will decrease. So, the second blank is "decrease".

Question 2) Two important concepts exist when it comes to chemical equilibrium: 

- In a chemical equilibrium, the forward and reverse reactions don't stop, they simply occur at the same rate or velocity.

- The observation above implies that the concentration of each participant in the reaction remains constant.

Hence, we can already rule out option 1. In option 3, since we are not dealing with any gaseous substance, we can rule it out, because we won't observe any change in pressure just due to the solubility of the copper II sulfate (all substances are either liquid, solid or aqueous).

If the mass of the copper II sulfate were completely dissolved, this means that there would be no more reverse reaction, and this is not the case in a chemical equilibrium, only in an irreversible reaction.

Our only option is option number 4, the mass of $CuSO_4$CuSO4 remaining is constant, because $CuSO_4$CuSO4 is being consumed at the same rate that is being formed, not changing its remaining mass. If the flask was stoppered, this means that the volume inside it is constant.