The reaction: 2A(g) + B(g) → 2C(g) was studied at 80 °C and the following results were obtained. What is the order of reaction with respect to: (α) A? (β) B? What is the overall order of reaction?

In Experiment 1, $[\text{A}] = 0.100 \text{ mol dm}^{-3}$, $[\text{B}] = 0.100 \text{ mol dm}^{-3}$, Rate $= 2.0 \times 10^{-4} \text{ mol dm}^{-3}\text{s}^{-1}$.  
In Experiment 2, $[\text{A}] = 0.200 \text{ mol dm}^{-3}$, $[\text{B}] = 0.100 \text{ mol dm}^{-3}$, Rate $= 4.0 \times 10^{-4} \text{ mol dm}^{-3}\text{s}^{-1}$.  
When $[\text{B}]$ is kept constant and $[\text{A}]$ is doubled (from 0.100 to 0.200), the rate doubles (from $2.0 \times 10^{-4}$ to $4.0 \times 10^{-4}$).  
This indicates that the reaction is first order with respect to A.  
The order of reaction with respect to A is $\boxed{\text{1}}$.

• (β) What is the order of reaction with respect to B?

In Experiment 2, $[\text{A}] = 0.200 \text{ mol dm}^{-3}$, $[\text{B}] = 0.100 \text{ mol dm}^{-3}$, Rate $= 4.0 \times 10^{-4} \text{ mol dm}^{-3}\text{s}^{-1}$.  
In Experiment 3, $[\text{A}] = 0.200 \text{ mol dm}^{-3}$, $[\text{B}] = 0.200 \text{ mol dm}^{-3}$, Rate $= 1.6 \times 10^{-3} \text{ mol dm}^{-3}\text{s}^{-1}$.  
When $[\text{A}]$ is kept constant and $[\text{B}]$ is doubled (from 0.100 to 0.200), the rate increases by a factor of $\frac{1.6 \times 10^{-3}}{4.0 \times 10^{-4}} = 4$.  
Since doubling $[\text{B}]$ quadruples the rate, the reaction is second order with respect to B.  
The order of reaction with respect to B is $\boxed{\text{2}}$.

(c)(ii)

• What is the overall order of reaction? The overall order of reaction is the sum of the individual orders with respect to each reactant. Overall order $= (\text{order with respect to A}) + (\text{order with respect to B}) = 1 + 2 = 3$. The overall order of reaction is $\boxed{\text{3}}$.

Question 5 (Thermodynamics & Hybridization)

(a)
Hess's law of constant heat summation states that the total enthalpy change for a chemical reaction is independent of the pathway taken, as long as the initial and final conditions are the same. It means that if a reaction can be expressed as a sum of several steps, the enthalpy change for the overall reaction is the sum of the enthalpy changes for each step.

(b)
The thermochemical equations for the given statements are:

• (i) Standard enthalpy of formation of LiF$_{(s)}$ is -617 kJ mol$^{-1}$. $$\text{Li}_{(s)} + \frac{1}{2}\text{F}_{2(g)} \longrightarrow \text{LiF}_{(s)} \quad \Delta H_f^\circ = -617 \text{ kJ mol}^{-1}$$
• (ii) Atomization energy of Li$_{(s)}$ is +161 kJ mol$^{-1}$. $$\text{Li}_{(s)} \longrightarrow \text{Li}_{(g)} \quad \Delta H_{atom} = +161 \text{ kJ mol}^{-1}$$
• (iii) Ionisation energy of Li$_{(g)}$ is +520 kJ mol$^{-1}$. $$\text{Li}_{(g)} \longrightarrow \text{Li}^{+}_{(g)} + \text{e}^{-} \quad \Delta H_{ion} = +520 \text{ kJ mol}^{-1}$$
• (iv) Dissociation energy of F$_{2(g)}$ is +79 kJ mol$^{-1}$. $$\frac{1}{2}\text{F}_{2(g)} \longrightarrow \text{F}_{(g)} \quad \Delta H_{diss} = +\frac{79}{2} = +39.5 \text{ kJ mol}^{-1}$$
• (v) Electron affinity of F$_{(g)}$ is -328 kJ mol$^{-1}$. $$\text{F}_{(g)} + \text{e}^{-} \longrightarrow \text{F}^{-}_{(g)} \quad \Delta H_{EA} = -328 \text{ kJ mol}^{-1}$$

(c)(i)

• Use the thermochemical equations in (b) to draw an energy cycle. The Born-Haber cycle for the formation of LiF$_{(s)}$ is shown below: $$\begin{array}{c} \text{Li}_{(s)} + \frac{1}{2}\text{F}_{2(g)} \\ \downarrow \Delta H_{atom}(\text{Li}) = +161 \\ \text{Li}_{(g)} + \frac{1}{2}\text{F}_{2(g)} \\ \downarrow \Delta H_{ion}(\text{Li}) = +520 \\ \text{Li}^{+}_{(g)} + \text{e}^{-} + \frac{1}{2}\text{F}_{2(g)} \\ \downarrow \Delta H_{diss}(\text{F}_2) = +39.5 \\ \text{Li}^{+}_{(g)} + \text{e}^{-} + \text{F}_{(g)} \\ \downarrow \Delta H_{EA}(\text{F}) = -328 \\ \text{Li}^{+}_{(g)} + \text{F}^{-}_{(g)} \\ \downarrow \Delta H_{lattice} \\ \text{LiF}_{(s)} \\ \uparrow \Delta H_f^\circ = -617 \\ \text{Li}_{(s)} + \frac{1}{2}\text{F}_{2(g)} \end{array}$$ (Note: The arrows for $\Delta H_f^\circ$ and $\Delta H_{lattice}$ should point downwards, indicating exothermic processes for formation and lattice energy, respectively, as drawn in the cycle above. The cycle starts from elements in standard states, goes through gaseous ions, and then forms the ionic solid.)

(c)(ii)

• Using Hess's law, write an equation relating all the energy terms in (b). According to Hess's Law, the enthalpy of formation of LiF is equal to the sum of the enthalpy changes in the Born-Haber cycle: $$\Delta H_f^\circ = \Delta H_{atom}(\text{Li}) + \Delta H_{ion}(\text{Li}) + \Delta H_{diss}(\text{F}_2) + \Delta H_{EA}(\text{F}) + \Delta H_{lattice}$$

(c)(iii)

• Using the equation in (ii) calculate the lattice energy of LiF$_{(s)}$.

$$
\Delta H_{lattice} = -617 - 392.5
$$
$$
\Delta H_{lattice} = -1009.5 \text{ kJ mol}^{-1}
$$
The lattice energy of LiF$_{(s)}$ is $\boxed{\text{-1009.5 kJ mol}^{-1}\text{}}$.

(d)(i)

• Give the type of hybridization shown by the central atoms in each of the following compounds:
  • BCl$_{3}$: The central atom is Boron (B). Boron has 3 valence electrons and forms 3 single bonds with chlorine atoms. There are no lone pairs on boron. The steric number is 3 (3 bonding pairs + 0 lone pairs), which corresponds to $\boxed{\text{sp}^{2}\text{}}$ hybridization.
  • NH$_{3}$: The central atom is Nitrogen (N). Nitrogen has 5 valence electrons and forms 3 single bonds with hydrogen atoms, leaving 1 lone pair. The steric number is 4 (3 bonding pairs + 1 lone pair), which corresponds to $\boxed{\text{sp}^{3}\text{}}$ hybridization.
  • H$_{2}$O: The central atom is Oxygen (O). Oxygen has 6 valence electrons and forms 2 single bonds with hydrogen atoms, leaving 2 lone pairs. The steric number is 4 (2 bonding pairs + 2 lone pairs), which corresponds to $\boxed{\text{sp}^{3}\text{}}$ hybridization.

(d)(ii)

• Deduce the shape of each of the molecules in (i):
  • BCl$_{3}$: With 3 bonding pairs and 0 lone pairs around the central boron atom, the electron geometry and molecular geometry are both trigonal planar. The shape is $\boxed{\text{trigonal planar}}$.
  • NH$_{3}$: With 3 bonding pairs and 1 lone pair around the central nitrogen atom, the electron geometry is tetrahedral, but the molecular geometry is trigonal pyramidal due to the repulsion from the lone pair. The shape is $\boxed{\text{trigonal pyramidal}}$.
  • H$_{2}$O: With 2 bonding pairs and 2 lone pairs around the central oxygen atom, the electron geometry is tetrahedral, but the molecular geometry is bent (or V-shaped) due to the repulsion from the two lone pairs. The shape is $\boxed{\text{bent}}$.