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Hydrogen and iodine are sealed in a 2 dm³ container - NSC Physical Sciences - Question 6 - 2017 - Paper 2
Question 6
Hydrogen and iodine are sealed in a 2 dm³ container. The reaction is allowed to reach equilibrium at 700 K according to the following balanced equation: H₂(g) + I₂(... show full transcript
Worked Solution & Example Answer:Hydrogen and iodine are sealed in a 2 dm³ container - NSC Physical Sciences - Question 6 - 2017 - Paper 2
Step 1
Give a reason why changes in pressure will have no effect on the equilibrium position.
Answer
Changes in pressure will not affect the equilibrium position for a gas-phase reaction when the number of moles of gaseous reactants is equal to the number of moles of gaseous products. In this case, the reaction involves 1 mole of H₂ and 1 mole of I₂ producing 2 moles of HI, thus leading to no change in the overall pressure equilibrium.
Step 2
Calculate the initial mass of I₂(g), in grams, that was sealed in the container, if Kc for the reaction is 55.3 at 700 K.
Answer
To find the initial mass of I₂(g), we start with the equilibrium concentrations:
-
Calculate the initial concentration of H₂ and I₂ at equilibrium:
- For H₂:
- Mole concentration at equilibrium: 0.028 mol in 2 dm³ => [H_2] = rac{0.028}{2} = 0.014 ext{ mol dm}^{-3}
- For I₂:
- Mole concentration at equilibrium: 0.017 mol in 2 dm³ => [I_2] = rac{0.017}{2} = 0.0085 ext{ mol dm}^{-3}
- For H₂:
-
Use the equilibrium constant expression: K_c = rac{[HI]^2}{[H_2][I_2]} Given that Kc = 55.3,
- We can rearrange it to find [HI]: 55.3 = rac{[HI]^2}{0.014 imes 0.0085}
-
Solve for [HI]:
ightarrow [HI] ≈ 0.08112 ext{ mol dm}^{-3}$$
-
Determine the amount of I₂ that reacted:
- Initial I₂ = x (initial amount at the start)
- Change at equilibrium = -0.017 (because it’s a reactant)
- Remaining I₂ at equilibrium = x - 0.017
- At equilibrium, we already calculated [I₂] = 0.0085. Therefore:
-
Calculate the initial mass of I₂ sealed in grams:
- Molar mass of I₂ = 254 g/mol, .
Step 3
What do the parallel lines in the first two minutes indicate?
Answer
The parallel lines in the first two minutes indicate that the reaction rates of the forward and reverse reactions are equal and that equilibrium is being established. There is no change in the concentration of reactants or products during this time period.
Step 4
State TWO possible changes that could be made to the reaction conditions at t = 2 minutes.
Answer
- Increase the temperature of the system, which could favor the endothermic direction of the reaction.
- Decrease the volume of the container, thereby increasing pressure, which could favor the side of the reaction with fewer moles of gas.
Step 5
Is the forward reaction EXOTHERMIC or ENDOTHERMIC? Fully explain the answer.
Answer
The forward reaction is Exothermic. This is because, as the reaction progresses to produce HI, it releases energy as heat. According to Le Chatelier's principle, any change in conditions affecting the system at equilibrium will favor the direction that counteracts that change, which indicates that forming products releases heat.
Step 6
How will this change influence the Kc value? Choose from INCREASES, DECREASES or REMAINS THE SAME.
Answer
The Kc value will DECREASE if the temperature is increased, since for an exothermic reaction, increasing temperature shifts the equilibrium to favor the reactants, thereby reducing the concentrations of products, which in turn lowers the Kc value.
Step 7
What change was made to the equilibrium mixture at t = 8 minutes?
Answer
At t = 8 minutes, a potential addition or removal of reactant or product could have occurred, leading to a shift in the reaction rate as depicted on the graph. This change might involve adjusting concentration or changing the physical conditions of the reactants.