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'IN EXCESS' means that there is more than enough of the reactant (nitric acid) to react completely with the other reactant (copper). The reactant in excess will not be fully consumed.

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5.1. 'IN EXCESS' means that there is more than enough of the reactant (nitric acid) to react completely with the other reactant (copper). The reactant in excess will not be fully consumed.

5.2.1. Between t3\text{t}_3 and t4\text{t}_4, the concentration of reactants (copper and nitric acid) decreases as they are consumed. According to collision theory, a decrease in reactant concentration leads to a decrease in the frequency of effective collisions between reactant particles. This results in a decrease in the rate of reaction.

5.2.2. The rate of the reaction at t2\text{t}_2 compared to that at t1\text{t}_1 is HIGHER THAN.

5.2.3. The slope of the graph (volume of NO2\text{NO}_2 vs. time) represents the rate of reaction. The curve is steeper at t2\text{t}_2 than at t1\text{t}_1, indicating a faster rate of production of NO2\text{NO}_2 and thus a faster reaction rate.

5.2.4.

  1. Use the formula for average rate.

    Average rate=ΔVolume of NO2ΔTime\text{Average rate} = \frac{\Delta \text{Volume of NO}_2}{\Delta \text{Time}}

  2. Substitute the given values and solve for t4\text{t}_4.

    0.02 dm3⋅s−1=1.7 dm3−0 dm3t4−0 s0.02 \text{ dm}^3\cdot\text{s}^{-1} = \frac{1.7 \text{ dm}^3 - 0 \text{ dm}^3}{\text{t}_4 - 0 \text{ s}} t4=1.70.02\text{t}_4 = \frac{1.7}{0.02} t4=85 s\text{t}_4 = \text{85 s}

    The value of time t4\text{t}_4 is 85 s\boxed{\text{85 s}}.

    5.2.5.
    The balanced chemical equation is:

    Cu(s)+4HNO3(conc.)→Cu(NO3)2(aq)+2NO2(g)+2H2O(l)\text{Cu(s)} + 4\text{HNO}_3\text{(conc.)} \to \text{Cu(NO}_3)_2\text{(aq)} + 2\text{NO}_2\text{(g)} + 2\text{H}_2\text{O(l)}

  3. Calculate the moles of NO2\text{NO}_2 produced at completion from the graph.

    From the graph, the maximum volume of NO2\text{NO}_2 produced is 1.92 dm3\text{dm}^3.

    Moles of NO2=VolumeMolar gas volume=1.92 dm325 dm3⋅mol−1=0.0768 mol\text{Moles of NO}_2 = \frac{\text{Volume}}{\text{Molar gas volume}} = \frac{1.92 \text{ dm}^3}{25 \text{ dm}^3\cdot\text{mol}^{-1}} = 0.0768 \text{ mol}

  4. Calculate the moles of HNO3\text{HNO}_3 that reacted using stoichiometry.

    From the equation, 4 moles of HNO3\text{HNO}_3 react for every 2 moles of NO2\text{NO}_2 produced.

    Moles of HNO3 reacted=0.0768 mol NO2×4 mol HNO32 mol NO2=0.1536 mol\text{Moles of HNO}_3 \text{ reacted} = 0.0768 \text{ mol NO}_2 \times \frac{4 \text{ mol HNO}_3}{2 \text{ mol NO}_2} = 0.1536 \text{ mol}

  5. Calculate the initial moles of HNO3\text{HNO}_3.

    Initial concentration of HNO3=12 mol⋅dm−3\text{HNO}_3 = 12 \text{ mol}\cdot\text{dm}^{-3}
    Volume of HNO3=20 cm3=0.020 dm3\text{HNO}_3 = 20 \text{ cm}^3 = 0.020 \text{ dm}^3

    Initial moles of HNO3=Concentration×Volume=12 mol⋅dm−3×0.020 dm3=0.24 mol\text{Initial moles of HNO}_3 = \text{Concentration} \times \text{Volume} = 12 \text{ mol}\cdot\text{dm}^{-3} \times 0.020 \text{ dm}^3 = 0.24 \text{ mol}

  6. Calculate the moles of HNO3\text{HNO}_3 remaining after the reaction.

    Moles of HNO3 remaining=Initial moles−Moles reacted\text{Moles of HNO}_3 \text{ remaining} = \text{Initial moles} - \text{Moles reacted} Moles of HNO3 remaining=0.24 mol−0.1536 mol=0.0864 mol\text{Moles of HNO}_3 \text{ remaining} = 0.24 \text{ mol} - 0.1536 \text{ mol} = 0.0864 \text{ mol}

  7. Calculate the concentration of HNO3\text{HNO}_3 remaining.

    The volume of the acid does not change, so it remains 0.020 dm3\text{dm}^3.

    Concentration of HNO3 remaining=Moles remainingVolume=0.0864 mol0.020 dm3=4.32 mol⋅dm−3\text{Concentration of HNO}_3 \text{ remaining} = \frac{\text{Moles remaining}}{\text{Volume}} = \frac{0.0864 \text{ mol}}{0.020 \text{ dm}^3} = 4.32 \text{ mol}\cdot\text{dm}^{-3}

    Since HNO3\text{HNO}_3 is a strong acid, it dissociates completely: HNO3→H++NO3−\text{HNO}_3 \to \text{H}^+ + \text{NO}_3^-.
    Therefore, the hydronium ion concentration is equal to the concentration of remaining HNO3\text{HNO}_3.

    Hydronium ion concentration=4.32 mol⋅dm−3\text{Hydronium ion concentration} = \boxed{4.32 \text{ mol}\cdot\text{dm}^{-3}}

    5.2.6. The reaction states that copper is oxidized in excess nitric acid, meaning copper is the limiting reactant. To produce more NO2\text{NO}_2 gas, the amount of the limiting reactant (copper) must be increased.
    A: Using the same mass of pure copper will produce the same amount of NO2\text{NO}_2.
    B: Using impure copper with greater mass implies a greater overall amount of material, which would likely contain a greater amount of actual copper, thus increasing the limiting reactant and producing more NO2\text{NO}_2.
    C: Increasing the concentration of HNO3\text{HNO}_3 (which is already in excess) will not increase the total amount of product

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