Question

3. 30.0cm3 of 2.00 moldm-3 HCl was placed in a conical flask and the temperature recorded as 21.0°C. When 0.0200 mol of potassium carbonate, K2CO3, was added to the acid and the mixture stirred with thermometer, the maximum temperature recorded was 26.2°C.
Calculate the enthalpy change of the neutralisation reaction.

Answer 1

Answer:

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Answer 2

Given:

30.0$cm^{3}$ of 2.00$mol dm^{-3}$ $HCl$ was placed in a conical flask and the temperature recorded as 21.0°C. When 0.0200 $mol$ of potassium carbonate, $K_{2}CO_{3}$, was added to the acid and the mixture stirred with thermometer, the maximum temperature recorded was 26.2°C.

To Find:

Enthalpy for this reaction

Solution:

Equation for this Reaction

$K_{2}CO_{3} + 2HCl$ →  $2KCl + H_{2}O + CO_{2}$

First we calculate heat produced in this reaction

$q = mc$Δ$T$

q = 30 × 4.18 × 5.2

q = 652.08 $J$ per 0.0200 mole of $K_{2}CO_{3}$

Enthalpy for this reaction

0.0200 mole of $K_{2}CO_{3}$ = 652.08 $J$

1 mole of $K_{2}CO_{3}$ = $\dfrac{652.08}{0.200} = 32604 J$

Enthalpy = $- 32.6 kJMol^{-1}$