Thermodynamics And Thermochemistry Result Question 33
Passage
When $100$ mL of $1.0$ $M $ $HCl$ was mixed with $100$ mL of $1.0$ $ M $ $NaOH$ in an insulated beaker at constant pressure, a temperature increase of $5.7^{\circ} \mathrm{C}$ was measured for the beaker and its contents (Expt. 1). Because the enthalpy of neutralisation of a strong acid with a strong base is a constant $\left(-57.0 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)$, this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. 2), $100$ mL of $2.0$ M acetic acid $\left(K_a=2.0 \times 10^{-5}\right.$ ) was mixed with $100$ mL of $1.0$ $M$ $ NaOH$ (under identical conditions to Expt. 1) where a temperature rise of $5.6^{\circ} \mathrm{C}$ was measured.
Enthalpy of dissociation (in $\mathrm{kJ} $ $\mathrm{mol}^{-1}$ ) of acetic acid obtained from the Expt. 2 is
(a) $1.0$
(b) $10.0$
(c) $24.5$
(d) $51.4$
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Answer:
Correct Answer: 33. $(1 \mathrm{~kJ} / \mathrm{mol})$
Solution:
- Let $C$ $ \mathrm{JK}^{-1}$ be the heat capacity of calorimeter.
Mass of solution $=200 \mathrm{~mL} \times 1 \mathrm{~g} \mathrm{~mL}^{-1}=200 \mathrm{~g}$
Heat evolved in Expt. 1
$=57 \times 1000 \times 0.1(\mathrm{~mol})=5700 \mathrm{~J} $
$\Rightarrow 5700 \mathrm{~J} =(200 \times 4.2+C) \times 5.7 $
$\Rightarrow 1000 =200 \times 4.2+C \hspace {20mm} $ …..(i)
Let $x \mathrm{~kJ} / \mathrm{mol}$ is heat evolved in neutralisation of acetic acid.
$\Rightarrow x \times 1000 \times 0.10 =(200 \times 4.2+C) \times 5.6 $
$\Rightarrow \frac{x \times 100}{5.6} =200 \times 4.2+C \hspace {20mm} $ …..(ii)
From (i) and (ii) : $x=56 \mathrm{~kJ} / \mathrm{mol}$
$\Rightarrow$ Enthalpy of ionisation of acetic acid
$=-56-(-57)=1 \mathrm{~kJ} / \mathrm{mol}$
BETA
