Photo AI
Carbonyl bromide, COBr₂, decomposes into carbon monoxide and bromine according to the following balanced equation: COBr₂(g) ⇌ CO(g) + Br₂(g) ΔH > 0 Initially COBr₂(g) is sealed in a 2 dm³ container and heated to 73 °C - NSC Physical Sciences - Question 6 - 2017 - Paper 2
Question 6
Carbonyl bromide, COBr₂, decomposes into carbon monoxide and bromine according to the following balanced equation: COBr₂(g) ⇌ CO(g) + Br₂(g) ΔH > 0 Initially COBr... show full transcript
Worked Solution & Example Answer:Carbonyl bromide, COBr₂, decomposes into carbon monoxide and bromine according to the following balanced equation: COBr₂(g) ⇌ CO(g) + Br₂(g) ΔH > 0 Initially COBr₂(g) is sealed in a 2 dm³ container and heated to 73 °C - NSC Physical Sciences - Question 6 - 2017 - Paper 2
Step 1
6.1 Define chemical equilibrium.
Answer
Chemical equilibrium is the state in a chemical reaction when the rate of the forward reaction equals the rate of the reverse reaction. It can also be defined as the state in a chemical reaction when the concentrations of reactants and products remain constant.
Step 2
6.2.1 Equilibrium concentration of the COBr₂(g)
Answer
To find the equilibrium concentration of COBr₂, we first calculate the initial quantity of CO(g) produced and then use the equilibrium constant.
-
Initial quantity of CO(g): 1,12 mol
-
Change in quantity for COBr₂(g): at equilibrium, we can assume that every mole of CO produced corresponds to the consumption of half that amount of COBr₂ due to stoichiometry:
-
If 1,12 mol of CO is produced, then 0.04 mol of COBr₂ will remain:
-
Initial COBr₂ = 0,04 mol + 0,04 mol (from CO produced) = 0,08 mol
-
-
Equilibrium concentration of COBr₂:
-
Kc calculation: Kc = [CO][Br₂] / [COBr₂]
-
Given Kc = 0,19, we determine:
-
Concentration: [COBr₂] = 0,02 mol dm⁻³.
-
Step 3
6.2.2 Percentage of COBr₂ that decomposed at 73 °C
Answer
To calculate the percentage of COBr₂ that decomposed:
- Initial quantity of COBr₂ before decomposition, assuming full initial volume is 0,04 mol, while at equilibrium we have 0,02 mol remaining.
- Amount decomposed = Initial - Equilibrium = 0,04 mol - 0,02 mol = 0,02 mol.
- Percentage decomposed = (Quantity decomposed / Initial quantity) × 100 = (0,02 mol / 0,04 mol) × 100 = 50%.
Step 4
6.3 Which ONE of the following CORRECTLY describes the Kc value when equilibrium is reached at a lower temperature?
Answer
As we lower the temperature for an endothermic reaction, the reaction shifts left to produce more reactants, thus reducing the equilibrium constant Kc; therefore, Kc < 0,19.
Step 5
6.4 How will the number of moles of COBr₂(g) be affected?
Answer
When the pressure of the system is decreased by increasing the volume, the equilibrium will shift towards increasing the number of moles of gas. Since COBr₂ decomposes into two moles of gas (CO and Br₂), the number of moles of COBr₂ will DECREASE as the system favors the production of more moles of gas.