**Measuring Enthalpy Change with a Coffee Cup Calorimeter**

A coffee cup calorimeter is used to determine ΔH for the reaction

NaOH(aq) + HCl(aq) → H_2O(\ell) + NaCl(aq) ΔH = ?

When 250. mL of 1.00 M NaOH was added to 250. mL of 1.00 M HCl at 1 bar, the temperature of the solution increased from 23.4 °C to 30.4 °C. Use this information to determine ΔH and complete the thermochemical expression. Assume that the heat capacities of the coffee cups, the temperature probe, and the stirrer are negligible, that the solution has the same density and the same specific heat capacity as water, and that there is no change in volume of the solutions upon mixing.

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ΔH = – 59 kJ

**Strategy and Explanation** Use the definition of specific heat capacity [Equation 6.2′ (← p. 186)] to calculate q_P, the heat transfer for the constant-pressure conditions (1 bar). Because the density of the solution is assumed to be the same as for water and the total volume is 500. mL, the mass of solution is 500. g. Because the reaction system heats the solution, q_P is negative and

c = \frac{q}{m × ΔT} [6.2]

q_P = – c × m × ΔT

= – (4.184 J g^{-1} °C^{-1} ) (500. g) (30.4 °C – 23.4 °C) = – 1.46 × 10^4 J = – 14.6 kJ

This quantity of heat transfer does not correspond to the equation as written, however; instead it corresponds to consumption of

250. mL × \frac{1.00 mol}{1000 mL} = 0.250 mol HCl and 250. mL × \frac{1.00 mol}{1000 mL} = 0.250 mol NaOH

From the balanced equation, 1 mol HCl is required for each 1 mol NaOH. Therefore the reactants are in the stoichiometric ratio and neither reactant is a limiting reactant. Because the chemical equation involves 1 mol HCl, the heat transfer must be scaled in proportion to this quantity of HCl.

\frac{Δ H}{1 mol HCl} = \frac{- 14.6 kJ}{0.250 mol HCl}

Δ H = 1 mol HCl × \frac{- 14.6 kJ}{0.250 mol HCl} = – 59 kJ

(Note that because the reactants were in the stoichiometric ratio, the NaOH could also have been used in the preceding calculation.)

**Reasonable Answer Check** The temperature of the surroundings increased, so the reaction is exothermic and ΔH° must be negative. The temperature of 500.-g solution went up 7.0 °C, so the heat transfer was about (500 × 7 × 4) J = 14,000 J = 14 kJ. This corresponded to one-quarter mole of each reactant, so the heat transfer per mole must be about 4 × 14 kJ = 56 kJ. Therefore ΔH should be about – 56 kJ, which it is.

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