AP Chemistry FRQ

Question

Gas	Initial Concentration (M)
H2	0.030
I2	0.015
HI	?

Samples of three gases, H2(g) , I2(g) , and HI(g) , were combined in a rigid vessel. The initial concentrations of H2(g) and I2(g) are given in the table above.

(a) The original value of the reaction quotient, Qc , for the reaction of H2(g) and I2(g) to form HI(g) (before any reactions take place and before equilibrium is established), was 5.56 . On the following graph, plot the points representing the initial concentrations of all three gases. Label each point with the formula of the gas.

 

Equilibrium was established at a certain temperature according to the following chemical equation.

H2(g)+I2(g)⇄2HI(g)ΔH°rxn=−9.4kJ/molrxn;Kc=49

After equilibrium was established, the concentration of H2(g) was 0.020M .

(b) On the graph above, carefully draw three curves, one for each of the three gases, starting from the initial points you drew in part ( a ). The curves must show how the concentration of each of the three gases changed as equilibrium was established.

 

 

H2(g) , I2(g) , and HI(g) are at equilibrium at a different temperature in a different vessel.

(c) When the temperature in the vessel is decreased, does the equilibrium shift to the right, favoring the product, or to the left, favoring the reactants? Justify your answer.

 

(d) Does the value of Kc increase, decrease, or remain the same when the temperature is decreased? Justify your answer based on the expression for Kc and the concentrations of the product and reactants.

 

(e) In the following empty box, draw an appropriate number of each type of molecule to represent a possible new equilibrium at the lower temperature.

Answer 1

I hope the AP Chem Exam went well! If anyone is still interested in an answer:

a) First, we have to find the initial concentration of HI. Based on the given chemical equation H₂(g) + I₂(g) ⇄ 2HI(g), we get the reaction quotient expression:

Qc = [HI]² / [H₂] [I₂]

With Qc = 5.56, [H₂] = 0.030M, and [I₂] = 0.015M, we get [HI] = 0.050M.

b) We know that at equilibrium, [H₂] = 0.020M. Because H₂ and I₂ react in a 1:1 ratio, the decrease in [H₂] must be the same as the decrease in [I₂], as each H₂ molecule reacted with one I₂ molecule. As there are two molecules of HI produced for each molecule of H₂ reacting, the increase in [HI] must be twice as much as the decrease in [H₂]. Therefore, as [H₂] decreases by 0.010M at equilibrium (from 0.030M to 0.020M), [I₂] must also decrease by 0.010M to be 0.005M at equilibrium; and [HI] must increase by 2*0.010M to result in a concentration of 0.070M at equilibrium.

On the graph, the curves for [H₂] and [I₂] would curve down and plateau, and the curve for [HI] would curve up and plateau.

c) Although heat is technically not considered as a reactant or product in the chemical equation, we can think of it as such. As we're given that ΔH°rxn=−9.4kJ/mol < 0, this means H_final - H_initial < 0, which is indicative of an exothermic reaction. Now, if we think of the chemical equation as H₂(g) + I₂(g) ⇄ HI(g) + heat, we can see that if the temperature decreases, the reaction equilibrium would shift to the right, favoring the product.

d) The equilibrium expression is Kc = [HI]² / [H₂] [I₂]. As forming the product HI would be favored, as discussed in part c), the concentration of HI in the numerator of the expression would increase, resulting in a larger Kc value.