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Write the observations and the equation of the reactions which occur when concentrated sulphuric (VI) acid reacts with each of the following: b) Copper (II) sulphate crystals c) Copper metal followed

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  1. Write the observations and the equation of the reactions which occur when concentrated sulphuric (VI) acid reacts with each of the following: b) Copper (II) sulphate crystals Observation: The blue crystals turn white. Explanation: Concentrated sulphuric (VI) acid is a strong dehydrating agent and removes the water of crystallization from the hydrated copper (II) sulphate. Equation:

    CuSO4⋅5H2O(s)→conc. H2SO4CuSO4(s)+5H2O(l)\text{CuSO}_4 \cdot 5\text{H}_2\text{O(s)} \xrightarrow{\text{conc. H}_2\text{SO}_4} \text{CuSO}_4\text{(s)} + 5\text{H}_2\text{O(l)}

    c) Copper metal followed by Potassium Manganate (VII) solution

    • Reaction with Copper metal: Observation: The copper metal dissolves, forming a blue solution. A colorless gas with a pungent smell is evolved. Equation: Cu(s)+2H2SO4(conc)→heatCuSO4(aq)+SO2(g)+2H2O(l)\text{Cu(s)} + 2\text{H}_2\text{SO}_4\text{(conc)} \xrightarrow{\text{heat}} \text{CuSO}_4\text{(aq)} + \text{SO}_2\text{(g)} + 2\text{H}_2\text{O(l)}
    • Reaction of the gas with Potassium Manganate (VII) solution: Observation: The purple potassium manganate (VII) solution is decolorized (turns colorless). Equation: 5SO2(g)+2KMnO4(aq)+2H2O(l)→K2SO4(aq)+2MnSO4(aq)+2H2SO4(aq)5\text{SO}_2\text{(g)} + 2\text{KMnO}_4\text{(aq)} + 2\text{H}_2\text{O(l)} \rightarrow \text{K}_2\text{SO}_4\text{(aq)} + 2\text{MnSO}_4\text{(aq)} + 2\text{H}_2\text{SO}_4\text{(aq)}

    d) Potassium Nitrate
    Observation: Colorless fumes (of nitric acid) are produced.
    Equation:

    KNO3(s)+H2SO4(conc)→heatHNO3(g)+KHSO4(s)\text{KNO}_3\text{(s)} + \text{H}_2\text{SO}_4\text{(conc)} \xrightarrow{\text{heat}} \text{HNO}_3\text{(g)} + \text{KHSO}_4\text{(s)}

  2. a) State Charles' Law. Charles' Law states that for a fixed mass of gas at a constant pressure, the volume is directly proportional to its absolute temperature.

    b) A gas occupies 450cm3450\text{cm}^3 at 27∘C27^\circ\text{C}. What volume would the gas occupy at 177∘C177^\circ\text{C}, if its pressure remains constant?

  1. Convert temperatures to Kelvin.

    $$
T_1 = 27^\circ\text{C} + 273 = 300\text{ K}
$$
$$
T_2 = 177^\circ\text{C} + 273 = 450\text{ K}
$$

  2. Apply Charles' Law, V1T1=V2T2\frac{V_1}{T_1} = \frac{V_2}{T_2}.

    $$
V_2 = \frac{V_1 T_2}{T_1}
$$
$$
V_2 = \frac{450\text{ cm}^3 \times 450\text{ K}}{300\text{ K}}
$$
$$
V_2 = \frac{202500}{300}\text{ cm}^3
$$
$$
V_2 = 675\text{ cm}^3
$$
The new volume would be $\boxed{\text{675 cm}^3}$.

c) At a temperature of $57^\circ\text{C}$, Nitrogen gas occupies a volume of $750\text{cm}^3$. At what temperature, in degrees Celsius, will the gas occupy $100\text{ cm}^3$?

  3. Convert initial temperature to Kelvin.

    $$
T_1 = 57^\circ\text{C} + 273 = 330\text{ K}
$$

  4. Apply Charles' Law, V1T1=V2T2\frac{V_1}{T_1} = \frac{V_2}{T_2}.

    $$
T_2 = \frac{V_2 T_1}{V_1}
$$
$$
T_2 = \frac{100\text{ cm}^3 \times 330\text{ K}}{750\text{ cm}^3}
$$
$$
T_2 = \frac{33000}{750}\text{ K}
$$
$$
T_2 = 44\text{ K}
$$

  5. Convert the final temperature back to degrees Celsius.

    $$
T_2(\text{in }^\circ\text{C}) = 44\text{ K} - 273 = -229^\circ\text{C}
$$
The gas will occupy $100\text{ cm}^3$ at $\boxed{\text{-229}^\circ\text{C}}$.

    13. a) State Graham's Law of Diffusion. Graham's Law of Diffusion states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass (or density) at constant temperature and pressure.

    b) It takes 20 seconds for $200\text{cm}^3$ of Oxygen gas to diffuse across a porous plug. How long will it take an equal volume of Sulphur (IV) oxide to diffuse across the same plug? (S = 32.0, O = 16.0)

  6. Calculate the molar masses of Oxygen (O2\text{O}_2) and Sulphur (IV) oxide (SO2\text{SO}_2).

    $$
M_{\text{O}_2} = 2 \times 16.0 = 32.0\text{ g/mol}
$$
$$
M_{\text{SO}_2} = 32.0 + (2 \times 16.0) = 32.0 + 32.0 = 64.0\text{ g/mol}
$$

  7. Apply Graham's Law of Diffusion.

    For equal volumes, the ratio of times is directly proportional to the square root of the ratio of molar masses. TimeO2TimeSO2=MO2MSO2 \frac{\text{Time}_{\text{O}_2}}{\text{Time}_{\text{SO}_2}} = \sqrt{\frac{M_{\text{O}_2}}{M_{\text{SO}_2}}}

  8. Substitute the given values.

    $$
\frac{20\text{ s}}{\text{Time}_{\text{SO}_2}} = \sqrt{\frac{32.0\text{ g/mol}}{64.0\text{ g/mol}}}
$$
$$
\frac{20\text{ s}}{\text{Time}_{\text{SO}_2}} = \sqrt{0.5}
$$
$$
\frac{20\text{ s}}{\text{Time}_{\text{SO}_2}} \approx 0.7071
$$

  9. Solve for TimeSO2\text{Time}_{\text{SO}_2}.

    $$
\text{Time}_{\text{SO}_2} = \frac{20\text{ s}}{0.7071}
$$
$$
\text{Time}_{\text{SO}_2} \approx 28.28\text{ s}
$$
It will take approximately $\boxed{\text{28.28 s}}$ for Sulphur (IV) oxide to diffuse.
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