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Explain the term chemical equilibrium. At which times interval(s) is / (are) the reaction at equilibrium? Does the FORWARD or the REVERSE reaction have the faster reaction rate during the interval t2 to t3?

Chemistry

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1.1 Chemical equilibrium is a state in a reversible reaction where the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of reactants and products remain constant.

Part 1.2

The reaction is at equilibrium during the time intervals where the concentrations of all reactants and products are constant.

The concentrations are constant during the intervals:
t1 to t2 and t3 to t4\boxed{\text{t}_1 \text{ to t}_2 \text{ and t}_3 \text{ to t}_4}

1.3 During the interval t2t_2 to t3t_3, the concentration of CO2\text{CO}_2 (product) decreases, and the concentrations of CO\text{CO} and O2\text{O}_2 (reactants) increase. This indicates that the reverse reaction is favored.
REVERSE reaction\boxed{\text{REVERSE reaction}}

1.4 At t4t_4, the temperature is decreased. The graph shows that after t4t_4, the concentration of CO2\text{CO}_2 increases, and the concentrations of CO\text{CO} and O2\text{O}_2 decrease. This means the forward reaction is favored. Since a decrease in temperature favors the forward reaction, the forward reaction must be exothermic.
EXOTHERMIC\boxed{\text{EXOTHERMIC}}

1.5 When the temperature is decreased, the system shifts to favor the reaction that releases heat (exothermic reaction) to counteract the change. Since the forward reaction is favored (indicated by the increase in product CO2\text{CO}_2 and decrease in reactants CO\text{CO} and O2\text{O}_2), the forward reaction must be exothermic.

1.6 At t2t_2, the concentration of CO2\text{CO}_2 (product) decreases sharply, while the concentrations of CO\text{CO} and O2\text{O}_2 (reactants) increase sharply. This indicates that some of the product, CO2\text{CO}_2, was removed from the system.
Removal of CO2\boxed{\text{Removal of } \text{CO}_2}

Part 1.7

The value of KcK_c for a given reaction is only affected by temperature. Since the change made at t2t_2 (removal of CO2\text{CO}_2) is not a temperature change, the value of KcK_c will remain the same.

REMAIN THE SAME\boxed{\text{REMAIN THE SAME}}

Part 1.8· 5 steps

  1. Calculate the initial moles of CO\text{CO} and O2\text{O}_2.

    Molar mass of CO=12.01+16.00=28.01 g/mol\text{CO} = 12.01 + 16.00 = 28.01 \text{ g/mol}
    Molar mass of O2=2×16.00=32.00 g/mol\text{O}_2 = 2 \times 16.00 = 32.00 \text{ g/mol}
    Initial moles of CO=63 g28.01 g/mol=2.249 mol\text{CO} = \frac{63 \text{ g}}{28.01 \text{ g/mol}} = 2.249 \text{ mol}
    Initial moles of O2=9.11 g32.00 g/mol=0.2847 mol\text{O}_2 = \frac{9.11 \text{ g}}{32.00 \text{ g/mol}} = 0.2847 \text{ mol}

  2. Set up an ICE table using initial concentrations.

    Volume of container = 2 dm32 \text{ dm}^3
    Initial concentration of CO=2.249 mol2 dm3=1.1245 mol/dm3\text{CO} = \frac{2.249 \text{ mol}}{2 \text{ dm}^3} = 1.1245 \text{ mol/dm}^3
    Initial concentration of O2=0.2847 mol2 dm3=0.14235 mol/dm3\text{O}_2 = \frac{0.2847 \text{ mol}}{2 \text{ dm}^3} = 0.14235 \text{ mol/dm}^3
    Initial concentration of CO2=0 mol/dm3\text{CO}_2 = 0 \text{ mol/dm}^3

    The balanced equation is: 2CO(g)+O2(g)⇌2CO2(g)2\text{CO}(\text{g}) + \text{O}_2(\text{g}) \rightleftharpoons 2\text{CO}_2(\text{g})

    SpeciesInitial (mol/dm3^3)Change (mol/dm3^3)Equilibrium (mol/dm3^3)
    CO\text{CO}1.1245−2x-2x1.1245−2x1.1245 - 2x
    O2\text{O}_20.14235−x-x0.14235−x0.14235 - x
    CO2\text{CO}_20+2x+2x2x2x

  3. Use the given equilibrium concentration of CO2\text{CO}_2 to find xx.

    At equilibrium, [CO2]=0.15 mol/dm3[\text{CO}_2] = 0.15 \text{ mol/dm}^3.
    So, 2x=0.15 mol/dm32x = 0.15 \text{ mol/dm}^3
    x=0.152=0.075 mol/dm3x = \frac{0.15}{2} = 0.075 \text{ mol/dm}^3

  4. Calculate the equilibrium concentrations of CO\text{CO} and O2\text{O}_2.

    [CO]=1.1245−2x=1.1245−2(0.075)=1.1245−0.15=0.9745 mol/dm3[\text{CO}] = 1.1245 - 2x = 1.1245 - 2(0.075) = 1.1245 - 0.15 = 0.9745 \text{ mol/dm}^3
    [O2]=0.14235−x=0.14235−0.075=0.06735 mol/dm3[\text{O}_2] = 0.14235 - x = 0.14235 - 0.075 = 0.06735 \text{ mol/dm}^3

  5. Write the expression for KcK_c and substitute the equilibrium concentrations.

    Kc=[CO2]2[CO]2[O2]K_c = \frac{[\text{CO}_2]^2}{[\text{CO}]^2[\text{O}_2]} Kc=(0.15)2(0.9745)2(0.06735)K_c = \frac{(0.15)^2}{(0.9745)^2(0.06735)} Kc=0.0225(0.94965)(0.06735)K_c = \frac{0.0225}{(0.94965)(0.06735)} Kc=0.02250.06402K_c = \frac{0.0225}{0.06402} Kc=0.35145K_c = 0.35145

    Rounding to three significant figures:
    Kc=0.351\boxed{K_c = 0.351}

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