Equilibrium constant questions

In summary, the equilibrium constant for a reaction at one atmosphere in a closed container is not affected by the amount of solid present. Therefore, decreasing the volume of the container, adding more solid NaOH, or adding N2 gas will not decrease the total amount of CO2 gas present at equilibrium. The correct answer is none of the above. However, it is important to note that adding more solid NaOH can increase the rate of the reaction by providing more surface area for the reaction to occur on.
  • #1
ada0713
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Homework Statement



The following reaction is at equilibrium at one atmosphere, in a closed container
NaOH(s) + CO2(g)-> NaHCO3 (s)
which, if any, of the following actions will decrease the total amount of CO2 gas present at equilibrium?

1)decreasing the volume of the container
2) none of the above
3) removing half of the solid NaHCO3
4)adding mlore solid NaOH
5)adding N2 gas to double the pressure

The Attempt at a Solution



I chose (1) and got it wrong, so I chose (5) next but it was wrong again,
I thought that solid is not included in equilibrium constant which means that
it has little effect on equilibrium change, that's why I eliminated (3) and (4).
I guess for, (1), decreasing the volume wouldn't do much
to shift the equilibrium constant to the right.. then

I don't understand why (5) is wrong because if you add more gas,
it will double the pressure thus makeing the reactants to shift to the
right to re-establish the equilibrium constant,,

Is the answer none of the above?

Please help!
 
Last edited:
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  • #2
5 is wrong because the equilibrium concentration of CO2 in the gas phase is constant regardless of how much pressure there is from N2. The concentration of CO2 is actually the partial pressure of CO2 in that case. Your thinking on #3 is logical but not #4. Remember that the reaction occurs on the surface of the sodium hydroxide solid. What would adding more NaOH surface do?
 
  • #3
When you add more solid NaOH,, more CO2 molecules will react with NaOH; thus, decreasing the moles of CO2.. Am I right,,?
But still, does solid really have anything to do with this?

I don't think I understand the reason why we don't include solid in equilibrium constant
 

Related to Equilibrium constant questions

1. What is equilibrium constant?

Equilibrium constant, denoted as K, is a numerical value that expresses the relationship between the concentrations of reactants and products at equilibrium in a chemical reaction. It helps determine the direction and extent of a chemical reaction.

2. How is equilibrium constant calculated?

The equilibrium constant is calculated by dividing the concentration of products raised to their respective stoichiometric coefficients by the concentration of reactants raised to their respective stoichiometric coefficients. This can be represented as: K = [products]^m/[reactants]^n, where m and n are the stoichiometric coefficients of the products and reactants, respectively.

3. What does the value of equilibrium constant indicate?

The value of equilibrium constant indicates the relative amounts of reactants and products present at equilibrium. A large K value (greater than 1) indicates a high concentration of products at equilibrium, while a small K value (less than 1) indicates a high concentration of reactants at equilibrium.

4. Can the equilibrium constant be changed?

The equilibrium constant is a characteristic of a specific chemical reaction at a given temperature. It cannot be changed by altering the concentrations of reactants or products, but it can be altered by changing the temperature of the reaction.

5. How does temperature affect equilibrium constant?

The equilibrium constant is affected by temperature through the change in the value of the equilibrium constant expression. In general, an increase in temperature leads to an increase in the value of K for an endothermic reaction, and a decrease in the value of K for an exothermic reaction.

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