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Define the term rate of reaction.

Chemistry

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Answer

0.00585 dm3⋅s−1\text{0.00585 dm}^3 \cdot \text{s}^{-1}

Step-by-step solution

6 steps
  1. Define the term rate of reaction.

    The rate of reaction is the change in concentration of reactants or products per unit time.

  2. Identify a controlled variable.

    One controlled variable for this investigation is the mass/surface area of magnesium carbonate, temperature, or pressure.

  3. Explain why the mass decreases.

    The mass of the reaction mixture and flask decreases because carbon dioxide gas (CO2(g)\text{CO}_2(\text{g})) is produced during the reaction and escapes from the flask.

  4. Calculate the average rate for curve B.

    From the graph for curve B:
    Initial mass of flask + contents at t=0 st=0 \text{ s} is 144.50 g144.50 \text{ g}.
    Final mass of flask + contents at t=120 st=120 \text{ s} is 143.24 g143.24 \text{ g}.

    Mass of CO2\text{CO}_2 produced:

    Δm(CO2)=144.50 g−143.24 g=1.26 g\Delta m(\text{CO}_2) = 144.50 \text{ g} - 143.24 \text{ g} = 1.26 \text{ g}

    Molar mass of CO2\text{CO}_2:

    M(CO2)=12.01 g/mol+(2×16.00 g/mol)=44.01 g/molM(\text{CO}_2) = 12.01 \text{ g/mol} + (2 \times 16.00 \text{ g/mol}) = 44.01 \text{ g/mol}

    Moles of CO2\text{CO}_2 produced:

    n(CO2)=Δm(CO2)M(CO2)=1.26 g44.01 g/mol≈0.028629857 moln(\text{CO}_2) = \frac{\Delta m(\text{CO}_2)}{M(\text{CO}_2)} = \frac{1.26 \text{ g}}{44.01 \text{ g/mol}} \approx 0.028629857 \text{ mol}

    Volume of CO2\text{CO}_2 produced (molar gas volume is 24.5 dm3⋅mol−124.5 \text{ dm}^3 \cdot \text{mol}^{-1}):

    V(CO2)=n(CO2)×Molar gas volume=0.028629857 mol×24.5 dm3⋅mol−1≈0.7014315 dm3V(\text{CO}_2) = n(\text{CO}_2) \times \text{Molar gas volume} = 0.028629857 \text{ mol} \times 24.5 \text{ dm}^3 \cdot \text{mol}^{-1} \approx 0.7014315 \text{ dm}^3

    Average rate of CO2\text{CO}_2 production:

    Rate=V(CO2)Δt=0.7014315 dm3120 s≈0.005845 dm3⋅s−1\text{Rate} = \frac{V(\text{CO}_2)}{\Delta t} = \frac{0.7014315 \text{ dm}^3}{120 \text{ s}} \approx 0.005845 \text{ dm}^3 \cdot \text{s}^{-1}

    Rounding to three significant figures:
    The average rate is 0.00585 dm3⋅s−1\boxed{\text{0.00585 dm}^3 \cdot \text{s}^{-1}}.

  5. Identify the curve for Experiment 1 and explain using collision theory.

    Experiment 1 has the lowest HCl\text{HCl} concentration (0.1 mol⋅dm−30.1 \text{ mol} \cdot \text{dm}^{-3}). A lower concentration means a slower reaction rate, which corresponds to the curve with the least steep slope (slowest mass decrease).
    Therefore, curve A represents Experiment 1.
    According to collision theory, a lower concentration of HCl\text{HCl} means there are fewer HCl\text{HCl} particles per unit volume. This leads to a decrease in the frequency of effective collisions between HCl\text{HCl} and MgCO3\text{MgCO}_3 particles, resulting in a slower reaction rate.

  6. Compare the final mass of CO2\text{CO}_2 in Experiment 2 and Experiment 3.

    The final mass of CO2(g)\text{CO}_2(\text{g}) produced in Experiment 2 will be THE SAME as that in Experiment 3.
    This is because the problem states that hydrochloric acid is in excess. This means that magnesium carbonate (MgCO3\text{MgCO}_3) is the limiting reactant in both experiments. Since the initial amount of the limiting reactant (MgCO3\text{MgCO}_3) is the same for all experiments, the total amount of product (CO2\text{CO}_2) formed will be the same, regardless of the HCl\text{HCl} concentration.

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