A 'QwikCure' pack, used to treat sporting injuries, contains a bag of water inside a larger bag of finely powdered ammonium nitrate, NH4NO3 - VCE - SSCE Chemistry - Question 2 - 2009 - Paper 1

To sketch an energy profile diagram:

1. Label the y-axis as 'Energy (kJ mol$^{-1}$)'.
2. Mark an initial energy level for solid NH4NO3 at a lower point on the graph.
3. Draw a curved line that rises to a peak representing the transition state (activation energy).
4. The curve should then descend to a final energy level representing the dissolved NH4NO3 in aqueous form, which is above the initial level, reflecting the endothermic nature of the dissolution process.

The equilibrium reaction for ammonium ions (NH4+) with water (H2O) can be expressed as:

$\text{NH}_4^{+}(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{NH}_3(aq) + \text{H}_3\text{O}^{+}(aq)$

The most likely cause for the change at t2 is a decrease in the concentration of one of the reactants (either NH4+ or H2O) due to the removal of NH4+ ions or H3O+ ions. This decrease in reactant concentration leads to a lower rate of the forward reaction, which subsequently affects the back reaction as the chemical equilibrium is disrupted. The reduced concentration of products decreases the frequency of collisions, thus decreasing the rate of the back reaction.

The value of the equilibrium constant at time t2 would be equal to the value of the equilibrium constant at time t1, assuming the temperature has remained constant. This is because the equilibrium constant is only affected by changes in temperature, not concentrations of the reactants or products.

The acidity constant, Ka, for the reaction can be expressed as:

$K_a = \frac{[\text{NH}_3][\text{H}_3\text{O}^{+}]}{[\text{NH}_4^{+}]}$

To calculate the concentration of H3O+ ions:

1. The pH is given as 5.04.
2. Use the formula: $[H_3O^+] = 10^{-pH}$
3. Therefore:
   $[H_3O^+] = 10^{-5.04}$
4. This results in approximately $[H_3O^+] \approx 9.12 \times 10^{-6} \, mol \, L^{-1}$.

To find the mass of NH4NO3:

1. Use the molar concentration and volume of the solution: $n = [C] \times V$
2. Calculate moles: $n = 9.12 \times 10^{-6} imes 0.300 \, L = 2.736 \times 10^{-6} \text{ moles}$
3. The molar mass of NH4NO3 is approximately 80.04 g/mol.
4. Calculate the mass: $\text{mass} = n \times ext{molar mass} = 2.736 \times 10^{-6} \times 80.04 \approx 0.000218 g$.