Question

(a) describe how the percentages of carbon and hydrogen in an organic compound are determined. state the measurements which are made. 8 marks
(b) (i) state hesss law of constant heat summation
(ii) from the following enthalpy values, determine the amount of energy liberated when 1 mole of ethanol is completely burnt under standard conditions.
δhc° c(s) = -393.5 kj mol⁻¹
δhc° h₂(g) = -285.8 kj mol⁻¹
δhf° c₂h₅oh(l) = -227.6 kj mol⁻¹

Explanation:

Step1: Determine carbon and hydrogen percentages (a)

The organic - compound is burned completely in excess oxygen. Carbon is converted to carbon - dioxide and hydrogen to water. The mass of carbon - dioxide and water produced is measured.
The mass of carbon in the carbon - dioxide is calculated using the molar mass ratio. Since the molar mass of carbon is \(M_{C}=12\ g/mol\) and that of carbon - dioxide is \(M_{CO_{2}} = 44\ g/mol\), if the mass of \(CO_{2}\) is \(m_{CO_{2}}\), the mass of carbon \(m_{C}=\frac{12}{44}m_{CO_{2}}\).
The mass of hydrogen in water is calculated using the molar mass ratio. The molar mass of hydrogen is \(M_{H}=1\ g/mol\) and that of water is \(M_{H_{2}O}=18\ g/mol\). If the mass of \(H_{2}O\) is \(m_{H_{2}O}\), the mass of hydrogen \(m_{H}=\frac{2}{18}m_{H_{2}O}\).
The percentages of carbon and hydrogen are then calculated as \(\%C=\frac{m_{C}}{m_{compound}}\times100\) and \(\%H=\frac{m_{H}}{m_{compound}}\times100\), where \(m_{compound}\) is the mass of the organic compound.

Step2: State Hess's law (b)(i)

Hess's law of constant heat summation states that the enthalpy change of a chemical reaction is the same whether the reaction takes place in one step or in a series of steps, provided that the initial and final states are the same.

Step3: Calculate the enthalpy of combustion of ethanol (b)(ii)

The combustion of ethanol is \(C_{2}H_{5}OH(l)+3O_{2}(g)
ightarrow2CO_{2}(g)+3H_{2}O(l)\)
The enthalpy of formation of \(CO_{2}\) is \(\Delta H_{c}^{\circ}(C)= - 393.5\ kJ/mol\), for \(H_{2}O\) is \(\Delta H_{c}^{\circ}(H_{2})=-285.8\ kJ/mol\) and for \(C_{2}H_{5}OH\) is \(\Delta H_{f}^{\circ}(C_{2}H_{5}OH)=-227.6\ kJ/mol\)
\(\Delta H_{comb}^{\circ}(C_{2}H_{5}OH)=2\Delta H_{c}^{\circ}(C)+3\Delta H_{c}^{\circ}(H_{2})-\Delta H_{f}^{\circ}(C_{2}H_{5}OH)\)
\[

$$\begin{align*} \Delta H_{comb}^{\circ}(C_{2}H_{5}OH)&=2\times(- 393.5)+3\times(-285.8)-(-227.6)\\ &=-787-857.4 + 227.6\\ &=-1416.8\ kJ/mol \end{align*}$$

\]

Answer:

(a) Burn the compound in excess oxygen. Measure mass of \(CO_{2}\) and \(H_{2}O\) produced. Calculate mass of \(C\) and \(H\) using molar - mass ratios and then percentages.
(b)(i) The enthalpy change of a reaction is the same whether it occurs in one step or in a series of steps, given the same initial and final states.
(b)(ii) \(-1416.8\ kJ/mol\)