IBDP Chemistry -Reactivity 2.2 How fast? The rate of chemical change - IB Style Questions For SL Paper 2 -FA 2025

Detailed solution

(a)
One suitable method is to measure the volume of \(CO_2\) produced with time using a gas syringe (or a pressure sensor).
The rate can be found from \(\Delta(\text{volume of }CO_2)/\Delta t\) or from the gradient of a volume–time graph.

(b)
A catalyst provides an alternative reaction pathway with a lower activation energy, \(E_a\).
This means a greater fraction of particles have \(E \ge E_a\) at the same temperature, so there are more successful collisions per unit time and the rate increases.

(c)
Ester formed from methanoic acid and ethanol is ethyl methanoate.
Structural formula:
\(H{-}C(=O){-}O{-}CH_2{-}CH_3\)
Name: ethyl methanoate

(d)
(i)
\[ CH_3CH_2OH(l) + 3O_2(g) \rightarrow 2CO_2(g) + 3H_2O(g) \]
(ii)
Using average bond enthalpies (data booklet section 11):
Bond breaking (reactants)
In \(CH_3CH_2OH\): \(1\,C{-}C,\ 5\,C{-}H,\ 1\,C{-}O,\ 1\,O{-}H\)
In \(3O_2\): \(3\,O{=}O\)
\[ E_{\text{break}} = 1(346) + 5(414) + 1(358) + 1(463) + 3(498) \] \[ = 346 + 2070 + 358 + 463 + 1494 = 4731\ \text{kJ} \]
Bond forming (products)
In \(2CO_2\): \(4\,C{=}O\)
In \(3H_2O\): \(6\,O{-}H\)
\[ E_{\text{form}} = 4(804) + 6(463) = 3216 + 2778 = 5994\ \text{kJ} \]
\[ \Delta H \approx E_{\text{break}} – E_{\text{form}} = 4731 – 5994 = -1263\ \text{kJ mol}^{-1} \]
✅ \(\Delta H_{\text{comb}} \approx -1263\ kJ\,mol^{-1}\).