ESAT Chemistry Mock 4

First, compare the groups based on extraction methods. Metals extracted by electrolysis (Q and R) are more reactive than carbon. Metals extracted by reduction with carbon (S and T) are less reactive than carbon. Therefore, the group (Q, R) is more reactive than the group (S, T).

Second, compare S and T based on the reaction with acid. S reacts with dilute acid (placing it above hydrogen in the reactivity series), while T does not (placing it below hydrogen). Therefore, $\displaystyle S > T$ .

Third, compare Q and R using the displacement observation. A metal can only displace a less reactive metal from a solution of its salt. Since adding Q to a salt of R results in no reaction, Q cannot displace R. This implies that Q is less reactive than R. Therefore, $\displaystyle R > Q$ .

Combining these deductions gives the overall order: $\displaystyle R > Q > S > T$ .

We evaluate each statement based on the principles of fractional distillation and intermolecular forces.

* Statement 1: The boiling point range for diesel (250 – 350 °C) is higher than for petrol (40 – 200 °C). In a fractionating column, there is a temperature gradient, with it being hottest at the bottom and coolest at the top. Vapours rise and condense at different levels (trays) corresponding to their boiling points. Substances with higher boiling points, like diesel, condense at higher temperatures, which are found lower down the column. Therefore, statement 1 is correct.

* Statement 2: The boiling point range for kerosene (180 – 250 °C) is higher than for petrol (40 – 200 °C). Boiling involves overcoming the intermolecular forces between molecules. A higher boiling point means more energy is required to separate the molecules, which implies stronger intermolecular forces. Therefore, the intermolecular forces in kerosene are, on average, stronger than in petrol. Statement 2 is correct.

* Statement 3: The boiling point of alkanes (the main components of these fractions) increases with the length of the carbon chain. Longer chains have more electrons and a larger surface area, leading to stronger temporary dipole-dipole interactions (van der Waals forces). Since diesel has a much higher boiling point range than petrol, it must be composed of, on average, larger molecules with more carbon atoms. The statement claims the opposite. Therefore, statement 3 is incorrect.

Since only statements 1 and 2 are correct, the correct option is E.

We determine the number of protons ( $\displaystyle p$ ), neutrons ( $\displaystyle n$ ), and electrons ( $\displaystyle e$ ) for each species using their atomic numbers ( $\displaystyle \ce{O}$ : $\displaystyle Z = 8$ , $\displaystyle \ce{Ne}$ : $\displaystyle Z = 10$ , $\displaystyle \ce{Na}$ : $\displaystyle Z = 11$ ):

- For $\displaystyle \ce{^{18}O^{2-}}$ :
- $\displaystyle p = 8$ - $\displaystyle n = 18 - 8 = 10$ - $\displaystyle e = 8 - (-2) = 10$ - Electron configuration: 2,8
- Neutron-to-proton ratio: $\displaystyle \dfrac{10}{8} = 1.25$ - For $\displaystyle \ce{^{20}Ne+}$ :
- $\displaystyle p = 10$ - $\displaystyle n = 20 - 10 = 10$ - $\displaystyle e = 10 - (+1) = 9$ - Electron configuration: 2,7

- For $\displaystyle \ce{^{23}Na+}$ :
- $\displaystyle p = 11$ - $\displaystyle n = 23 - 11 = 12$ - $\displaystyle e = 11 - (+1) = 10$ - Electron configuration: 2,8
- Neutron-to-proton ratio: $\displaystyle \dfrac{12}{11} \approx 1.09$ Evaluating the statements:
1. Incorrect: $\displaystyle \ce{^{20}Ne+}$ has 9 electrons with configuration 2,7, not 2,8.
2. Correct: Both $\displaystyle \ce{^{18}O^{2-}}$ and $\displaystyle \ce{^{20}Ne+}$ have $\displaystyle 18 - 8 = 10$ and $\displaystyle 20 - 10 = 10$ neutrons, respectively.
3. Correct: The neutron-to-proton ratio in $\displaystyle \ce{^{18}O^{2-}}$ is $\displaystyle 1.25$ , which is greater than that in $\displaystyle \ce{^{23}Na+}$ ( $\displaystyle 1.09$ ).

Thus, statements 2 and 3 only are correct.

1. Effect of Temperature (Experiments 1 and 2):
Increasing the temperature from $\displaystyle 400\text{ K}$ to $\displaystyle 500\text{ K}$ at constant pressure decreases the equilibrium yield from $\displaystyle 35\%$ to $\displaystyle 18\%$ .
By Le Chatelier's principle, an increase in temperature shifts the equilibrium in the endothermic direction. Because the forward yield decreases, the forward reaction must be exothermic ( $\displaystyle \Delta H < 0$ ). This eliminates options B and D.

2. Effect of Pressure (Experiments 1 and 3):
Increasing the pressure from $\displaystyle 100\text{ kPa}$ to $\displaystyle 200\text{ kPa}$ at constant temperature increases the equilibrium yield from $\displaystyle 35\%$ to $\displaystyle 52\%$ .
By Le Chatelier's principle, an increase in pressure shifts the equilibrium towards the side with fewer moles of gas. Thus, the product side must have fewer gaseous moles than the reactant side ( $\displaystyle n_{\text{gas, products}} < n_{\text{gas, reactants}}$ ).
- In A: $\displaystyle 4\text{ mol gas} \to 2\text{ mol gas}$ ( $\displaystyle 2 < 4$ ), which matches.
- In C and F: $\displaystyle 2\text{ mol gas} \to 2\text{ mol gas}$ ( $\displaystyle \,\Delta n_{\text{gas}} = 0$ ), so pressure change would not affect the yield.
- In E: $\displaystyle 2\text{ mol gas} \to 3\text{ mol gas}$ , so pressure increase would decrease the yield.

3. Effect of Catalyst (Experiments 1 and 4):
Adding a catalyst reduces the time to reach equilibrium ( $\displaystyle 45\text{ min} \to 12\text{ min}$ ) without altering the equilibrium yield ( $\displaystyle 35\%$ ), which is consistent with the standard action of a catalyst.

Therefore, the only reaction consistent with all observations is A.

1. Calculate the energy absorbed by the solution: $\displaystyle \Delta T = 43.0^{\circ}\text{C} - 18.0^{\circ}\text{C} = 25.0^{\circ}\text{C}$ $\displaystyle q = m c \Delta T = 150\text{ g} \times 4.0\text{ J g}^{-1\circ}\text{C}^{-1} \times 25.0^{\circ}\text{C} = 15{,}000\text{ J} = 15.0\text{ kJ}$ 2. Determine the total energy that must be released by the reaction: $\displaystyle \text{Efficiency} = 75\% = 0.75$ $\displaystyle E_\text{released} = \dfrac{15.0\text{ kJ}}{0.75} = 20.0\text{ kJ}$ 3. Extract the enthalpy change from the reaction profile: $\displaystyle \Delta H = E_\text{products} - E_\text{reactants} = 100\text{ kJ mol}^{-1} - 300\text{ kJ mol}^{-1} = -200\text{ kJ mol}^{-1}$ Thus, $\displaystyle 1\text{ mol}$ of $\displaystyle \ce{Mg}$ releases $\displaystyle 200\text{ kJ}$ of energy.

4. Calculate the moles and mass of magnesium needed: $\displaystyle n(\ce{Mg}) = \dfrac{20.0\text{ kJ}}{200\text{ kJ mol}^{-1}} = 0.100\text{ mol}$ $\displaystyle \text{mass of } \ce{Mg} = 0.100\text{ mol} \times 24.0\text{ g mol}^{-1} = 2.40\text{ g}$

1. **Determine the charge on $\displaystyle \text{M}$ :**
In the oxide $\displaystyle \ce{M2O3}$ , the oxide ion is $\displaystyle \ce{O^{2-}}$ . Three oxide ions have a total charge of $\displaystyle 3 \times (-2) = -6$ . Therefore, the two $\displaystyle \text{M}$ ions must have a total charge of $\displaystyle +6$ , meaning each $\displaystyle \text{M}$ ion is $\displaystyle \ce{M^{3+}}$ .

2. **Sulfate of $\displaystyle \text{M}$ :**
The sulfate ion is $\displaystyle \ce{SO4^{2-}}$ . To balance the $\displaystyle +3$ charge of $\displaystyle \ce{M^{3+}}$ and the $\displaystyle -2$ charge of $\displaystyle \ce{SO4^{2-}}$ , the lowest common multiple is 6. We require two $\displaystyle \ce{M^{3+}}$ ions and three $\displaystyle \ce{SO4^{2-}}$ ions, giving the formula $\displaystyle \ce{M2(SO4)3}$ .

3. **Phosphate of $\displaystyle \text{M}$ :**
The phosphate ion is $\displaystyle \ce{PO4^{3-}}$ . Combining $\displaystyle \ce{M^{3+}}$ and $\displaystyle \ce{PO4^{3-}}$ gives a $\displaystyle 1:1$ ratio, resulting in the empirical formula $\displaystyle \ce{MPO4}$ .

4. **Nitride of $\displaystyle \text{M}$ :**
The nitride ion is $\displaystyle \ce{N^{3-}}$ (distinguished from the nitrate ion, $\displaystyle \ce{NO3-}$ ). Combining $\displaystyle \ce{M^{3+}}$ and $\displaystyle \ce{N^{3-}}$ gives a $\displaystyle 1:1$ ratio, resulting in the empirical formula $\displaystyle \ce{MN}$ .

Therefore, row E is the correct row.

1. Calculate total moles of $\displaystyle \ce{H+}$ supplied by the acids:
- From $\displaystyle \ce{H2SO4}$ : $\displaystyle n(\ce{H2SO4}) = 0.100\text{ dm}^3 \times 0.050\text{ mol dm}^{-3} = 0.0050\text{ mol}$ . Since $\displaystyle \ce{H2SO4}$ is diprotic, $\displaystyle n(\ce{H+})_{\ce{H2SO4}} = 2 \times 0.0050 = 0.0100\text{ mol}$ .
- From $\displaystyle \ce{HNO3}$ : $\displaystyle n(\ce{HNO3}) = 0.150\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 0.0150\text{ mol}$ of $\displaystyle \ce{H+}$ .

Total $\displaystyle n(\ce{H+}) = 0.0100 + 0.0150 = 0.0250\text{ mol}$ .

2. Calculate total moles of $\displaystyle \ce{OH-}$ supplied by the base:
- $\displaystyle n(\ce{Ba(OH)2}) = 0.250\text{ dm}^3 \times 0.040\text{ mol dm}^{-3} = 0.0100\text{ mol}$ .
- Since each formula unit of $\displaystyle \ce{Ba(OH)2}$ provides $\displaystyle 2\text{ OH-}$ ions, $\displaystyle n(\ce{OH-}) = 2 \times 0.0100 = 0.0200\text{ mol}$ .

3. Determine unreacted excess ions:
- Neutralization reaction: $\displaystyle \ce{H+ + OH- -> H2O}$ .
- $\displaystyle \ce{OH-}$ is limiting, so excess $\displaystyle n(\ce{H+}) = 0.0250 - 0.0200 = 0.0050\text{ mol}$ .

4. Calculate final concentration and pH:
- Total volume $\displaystyle V_{\text{total}} = 100 + 150 + 250 = 500\text{ cm}^3 = 0.500\text{ dm}^3$ .
- $\displaystyle [\ce{H+}] = \dfrac{0.0050\text{ mol}}{0.500\text{ dm}^3} = 0.010\text{ mol dm}^{-3} = 1.0 \times 10^{-2}\text{ mol dm}^{-3}$ .
- $\displaystyle \text{pH} = -\log_{10}[\ce{H+}] = -\log_{10}(1.0 \times 10^{-2}) = 2.0$ .

Hence, the correct option is B.

To find the bond enthalpy of $\displaystyle \ce{C\equiv N}$ , we use: $\displaystyle \Delta H = \sum (\text{bond enthalpies of bonds broken}) - \sum (\text{bond enthalpies of bonds formed})$ Bonds broken (reactants):
- $\displaystyle 4 \times \ce{C-H}$ bonds in $\displaystyle \ce{CH4} = 4 \times 415 = 1660\ \text{kJ}\ \text{mol}^{-1}$ - $\displaystyle 3 \times \ce{N-H}$ bonds in $\displaystyle \ce{NH3} = 3 \times 390 = 1170\ \text{kJ}\ \text{mol}^{-1}$ $\displaystyle \text{Total energy absorbed} = 1660 + 1170 = 2830\ \text{kJ}\ \text{mol}^{-1}$ Bonds formed (products):
- $\displaystyle 1 \times \ce{C-H}$ bond in $\displaystyle \ce{HCN} = 415\ \text{kJ}\ \text{mol}^{-1}$ - $\displaystyle 1 \times \ce{C\equiv N}$ bond in $\displaystyle \ce{HCN} = E(\ce{C\equiv N})$ - $\displaystyle 3 \times \ce{H-H}$ bonds in $\displaystyle 3\ce{H2} = 3 \times 435 = 1305\ \text{kJ}\ \text{mol}^{-1}$ $\displaystyle \text{Total energy released} = 415 + 1305 + E(\ce{C\equiv N}) = 1720 + E(\ce{C\equiv N})$ Setting up the equation: $\displaystyle +255 = 2830 - [1720 + E(\ce{C\equiv N})]$ $\displaystyle +255 = 1110 - E(\ce{C\equiv N})$ $\displaystyle E(\ce{C\equiv N}) = 1110 - 255 = 855\ \text{kJ}\ \text{mol}^{-1}$ Therefore, the bond enthalpy of the $\displaystyle \ce{C\equiv N}$ bond is $\displaystyle 855\ \text{kJ}\ \text{mol}^{-1}$ .

1. Cathode (reduction):
- Cations present in the aqueous solution: $\displaystyle \ce{Cu^{2+}}$ , $\displaystyle \ce{Na+}$ , $\displaystyle \ce{K+}$ , and $\displaystyle \ce{H+}$ (from the dissociation of water).
- In the reactivity series, potassium and sodium are far more reactive than hydrogen, so $\displaystyle \ce{K+}$ and $\displaystyle \ce{Na+}$ remain in solution.
- Copper is less reactive than hydrogen ( $\displaystyle E^\circ(\ce{Cu^{2+}/Cu}) = +0.34\text{ V} > E^\circ(\ce{H+/H2}) = 0.00\text{ V}$ ), so $\displaystyle \ce{Cu^{2+}}$ ions are reduced preferentially at the cathode: $\displaystyle \ce{Cu^{2+}(aq) + 2e- -> Cu(s)}$ 2. Anode (oxidation):
- Anions present in the aqueous solution: $\displaystyle \ce{SO4^{2-}}$ , $\displaystyle \ce{NO3-}$ , $\displaystyle \ce{Cl-}$ , and $\displaystyle \ce{OH-}$ (from water).
- Halide ions ( $\displaystyle \ce{Cl-}$ ) are discharged in preference to hydroxide ions and oxyanions (sulfate and nitrate do not oxidise under these conditions).
- Therefore, $\displaystyle \ce{Cl-}$ ions are oxidised at the anode: $\displaystyle 2\ce{Cl-(aq) -> Cl2(g) + 2e-}$ Combining both conclusions gives row A.

1. Determine the limiting reactant and volume of gas in Experiment 1:
- Moles of $\displaystyle \ce{Mg} = \dfrac{0.48\text{ g}}{24.0\text{ g mol}^{-1}} = 0.020\text{ mol}$ .
- Moles of $\displaystyle \ce{HCl} = 0.100\text{ dm}^3 \times 0.20\text{ mol dm}^{-3} = 0.020\text{ mol}$ .
- From the stoichiometry $\displaystyle \ce{Mg + 2HCl -> MgCl2 + H2}$ , $\displaystyle 0.020\text{ mol}$ of $\displaystyle \ce{HCl}$ reacts with only $\displaystyle 0.010\text{ mol}$ of $\displaystyle \ce{Mg}$ .
- Therefore, $\displaystyle \ce{HCl}$ is limiting in Experiment 1, and the amount of $\displaystyle \ce{H2}$ produced is $\displaystyle \dfrac{1}{2} \times 0.020\text{ mol} = 0.010\text{ mol}$ .
- On the graph, curve X reaches a plateau at $\displaystyle 2$ grid units, corresponding to $\displaystyle 0.010\text{ mol}$ of $\displaystyle \ce{H2}$ .

2. Determine the limiting reactant and volume of gas in Experiment 2:
- Moles of $\displaystyle \ce{Mg} = \dfrac{0.36\text{ g}}{24.0\text{ g mol}^{-1}} = 0.015\text{ mol}$ .
- Moles of $\displaystyle \ce{HCl} = 0.050\text{ dm}^3 \times 1.0\text{ mol dm}^{-3} = 0.050\text{ mol}$ .
- To react completely, $\displaystyle 0.015\text{ mol}$ of $\displaystyle \ce{Mg}$ requires $\displaystyle 2 \times 0.015 = 0.030\text{ mol}$ of $\displaystyle \ce{HCl}$ .
- Since $\displaystyle 0.050\text{ mol}$ of $\displaystyle \ce{HCl}$ is present, $\displaystyle \ce{HCl}$ is in excess and $\displaystyle \ce{Mg}$ is the limiting reactant.
- The amount of $\displaystyle \ce{H2}$ produced is equal to the moles of $\displaystyle \ce{Mg}$ reacted: $\displaystyle n(\ce{H2}) = 0.015\text{ mol}$ .
- The ratio of total volume produced in Experiment 2 to Experiment 1 is $\displaystyle \dfrac{0.015}{0.010} = 1.5$ .
- Therefore, the plateau height must be $\displaystyle 1.5 \times 2 = 3$ grid units.

3. Determine the initial rate of reaction:
- The initial rate depends on the concentration of $\displaystyle \ce{HCl(aq)}$ .
- In Experiment 2, $\displaystyle [\ce{HCl}] = 1.0\text{ mol dm}^{-3}$ , which is five times greater than in Experiment 1 ( $\displaystyle 0.20\text{ mol dm}^{-3}$ ).
- Hence, the initial rate is greater, meaning the curve has a steeper initial gradient than curve X.

Curve C has a steeper initial gradient than X and plateaus at $\displaystyle 3$ grid units.

To find the bond energy of the $\displaystyle \ce{C=O}$ bond, we use the relationship: $\displaystyle \Delta H = \sum (\text{bond energies of bonds broken}) - \sum (\text{bond energies of bonds formed})$ Bonds broken:
- In propanone ( $\displaystyle \ce{CH3COCH3}$ ): $\displaystyle 6 \times (\ce{C-H})$ , $\displaystyle 2 \times (\ce{C-C})$ , $\displaystyle 1 \times (\ce{C=O})$ - In $\displaystyle \ce{H2}$ : $\displaystyle 1 \times (\ce{H-H})$ $\displaystyle \text{Total broken} = 6(413) + 2(347) + 436 + E(\ce{C=O}) = 2478 + 694 + 436 + E(\ce{C=O}) = 3608 + E(\ce{C=O})$ Bonds formed:
- In propan-2-ol ( $\displaystyle \ce{CH3CH(OH)CH3}$ ): $\displaystyle 7 \times (\ce{C-H})$ , $\displaystyle 2 \times (\ce{C-C})$ , $\displaystyle 1 \times (\ce{C-O})$ , $\displaystyle 1 \times (\ce{O-H})$ $\displaystyle \text{Total formed} = 7(413) + 2(347) + 358 + 464 = 2891 + 694 + 358 + 464 = 4407\text{ kJ mol}^{-1}$ Alternatively, cancelling bonds unchanged on both sides ( $\displaystyle 6 \times \ce{C-H}$ and $\displaystyle 2 \times \ce{C-C}$ ): $\displaystyle \text{Net bonds broken} = E(\ce{C=O}) + E(\ce{H-H}) = E(\ce{C=O}) + 436$ $\displaystyle \text{Net bonds formed} = E(\ce{C-H}) + E(\ce{C-O}) + E(\ce{O-H}) = 413 + 358 + 464 = 1235$ Setting up the equation: $\displaystyle \Delta H = [E(\ce{C=O}) + 436] - 1235 = -54$ $\displaystyle E(\ce{C=O}) - 799 = -54$ $\displaystyle E(\ce{C=O}) = 799 - 54 = 745\text{ kJ mol}^{-1}$

Sulfur in the fuel can form $\displaystyle \ce{SO2}$ on combustion; $\displaystyle \ce{SO2}$ contributes to acid rain and environmental acidification.

Condensation changes gas to liquid: particles become closer and less free to move, and energy is released to the surroundings.

1. In the first run (Solvent 1), the solvent front moves $\displaystyle 12.0\text{ cm}$ :
- Distance moved by X along the first axis: $\displaystyle y_{\text{X}} = 0.65 \times 12.0\text{ cm} = 7.80\text{ cm}$ - Distance moved by Y along the first axis: $\displaystyle y_{\text{Y}} = 0.40 \times 12.0\text{ cm} = 4.80\text{ cm}$ - Separation along the first axis: $\displaystyle \Delta y = |7.80 - 4.80| = 3.00\text{ cm}$ 2. In the second run (Solvent 2), the solvent front moves $\displaystyle 10.0\text{ cm}$ perpendicular to the first direction:
- Distance moved by X along the second axis: $\displaystyle x_{\text{X}} = 0.20 \times 10.0\text{ cm} = 2.00\text{ cm}$ - Distance moved by Y along the second axis: $\displaystyle x_{\text{Y}} = 0.60 \times 10.0\text{ cm} = 6.00\text{ cm}$ - Separation along the second axis: $\displaystyle \Delta x = |6.00 - 2.00| = 4.00\text{ cm}$ 3. Since the two developments occur in perpendicular directions, the straight-line distance $\displaystyle d$ between the spots is given by Pythagoras' theorem: $\displaystyle d = \sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{(4.00\text{ cm})^2 + (3.00\text{ cm})^2} = \sqrt{16.00 + 9.00} = \sqrt{25.00} = 5.0\text{ cm}$ Hence, the correct option is D.