Chemistry · General chemistry I · Worked example

Find ΔH with a coffee-cup calorimeter

50.0 mL of 2.00 M HCl and 50.0 mL of 2.00 M NaOH, both at 21.0 °C, are mixed in a coffee-cup calorimeter, and the temperature rises to 34.3 °C. Find ΔH per mole of water formed. Treat the solution as water: 1.00 g/mL and 4.184 J/(g·°C).

q_{rxn} = - q_{soln}

Write the reaction

Hydrochloric acid and sodium hydroxide neutralize each other, forming water. They react one to one.

HCl (a q) + NaOH (a q) \to NaCl (a q) + H_{2} O (l)

\begin{array}{c} HCl (a q) + NaOH (a q) \\ \to NaCl (a q) + H_{2} O (l) \end{array}

Find the heat the solution gained

The mixture is 100.0 mL, so about 100.0 g, and it warms by 34.3 − 21.0 = 13.3 °C.

q_{soln} = (100.0 g) (4.184 J/ (g \cdot \circ C)) (13.3 \circ C) = 5.56 \times 10^{3} J

\begin{array}{c} q_{soln} = (100.0) (4.184) (13.3) J \\ = 5.56 \times 10^{3} J \end{array}

Find the heat the reaction released

The reaction supplied the heat the solution gained, so its q has the opposite sign.

q_{rxn} = - q_{soln} = - 5.56 kJ

Find the moles that reacted

Each solution holds 0.0500 L × 2.00 mol/L = 0.100 mol, and they react one to one, so 0.100 mol of water forms.

n = (0.0500 L) (2.00 mol/L) = 0.100 mol

\begin{array}{c} n = (0.0500 L) (2.00 mol/L) \\ = 0.100 mol \end{array}

Divide by the moles

ΔH is negative: the reaction is exothermic, releasing heat that warms the solution.

Δ H = \frac{- 5.56 kJ}{0.100 mol} = - 55.6 kJ/mol

Result

ΔH = −55.6 kJ per mole of water formed.

Your turn

In another trial, 100.0 mL of 1.00 M HCl and 100.0 mL of 1.00 M NaOH start at 20.0 °C and reach 26.6 °C. Find ΔH per mole of water formed.

Show the answer and explanation

−55 kJ/mol, to two significant figures.

The 200.0 g of solution gains (200.0 g)(4.184 J/(g·°C))(6.6 °C) = 5.5 × 10³ J, and 0.100 mol of water forms. ΔH = −5.5 kJ ÷ 0.100 mol = −55 kJ/mol. A rise of 6.6 °C has only two significant figures, so the answer does too.

Δ H = \frac{- 5.5 kJ}{0.100 mol} = - 55 kJ/mol

Keep exploring

In Calorimetry & heating curves, double the solution to 200.0 g with the same 13.3 °C rise: q doubles to 11.1 kJ, but so do the moles, so ΔH per mole stays −55.6 kJ/mol.

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