Question

In a constant-pressure calorimeter, 65.0 mL of 0.870 M H2SO4 was added to 65.0 mL of 0.450 M NaOH. The reaction caused the temperature of the solution to rise from 24.09 °C to 27.16 °C. If the solution has the same density and specific heat as water, what is ΔH for this reaction (per mole of H2O produced)? Assume that the total volume is the sum of the individual volumes.

Answer 1

Answer on Question #55347 – Chemistry – General chemistry

Question:

In a constant-pressure calorimeter, 65.0 mL of 0.870 M H₂SO₄ was added to 65.0 mL of 0.450 M NaOH. The reaction caused the temperature of the solution to rise from 24.09 °C to 27.16 °C. If the solution has the same density and specific heat as water, what is ΔH for this reaction (per mole of H₂O produced)? Assume that the total volume is the sum of the individual volumes.

Answer:

Conducted reaction is neutralization:

$$\mathrm{H_2SO_4} + 2\mathrm{NaOH} \rightarrow \mathrm{Na_2SO_4} + 2\mathrm{H_2O}$$

The amount of produced water is defined:

$$v(\mathrm{H_2O}) = v(\mathrm{NaOH}) = V(\mathrm{NaOH}) \times C(\mathrm{NaOH}) = 65\ \mathrm{ml} \times 0.450\ \mathrm{mol/L} = 29.25\ \mathrm{mmol} = 29.25 \times 10^{-3}\ \mathrm{mol}$$

The total mass of the solution is: $m = V\rho$, where $V$ – the total volume ($V = 65\ \mathrm{ml} + 65\ \mathrm{ml} = 130\ \mathrm{ml}$) and $\rho$ – the density of water ($\rho = 1\ \mathrm{g/ml}$).

$$m = 130\ \mathrm{ml} \times 1\ \mathrm{g/ml} = 130\ \mathrm{g}$$

Then find the heat released by this reaction:

$Q = mC\Delta t$, where $C$ – the specific heat for water which equals $4.2\ \mathrm{J\ g^{-1}\ ^{\circ}C^{-1}}$ and $\Delta t$ – the change of temperature ($\Delta t = 27.16\ ^{\circ}\mathrm{C} - 24.09\ ^{\circ}\mathrm{C} = 3.07\ ^{\circ}\mathrm{C}$)

$$Q = 130\ \mathrm{g} \times 4.2\ \mathrm{J\ g^{-1}\ ^{\circ}C^{-1}} \times 3.07\ ^{\circ}\mathrm{C} = 1676.22\ \mathrm{J}$$

Thus, $\Delta H = Q / v(\mathrm{H_2O}) = 1676.22\ \mathrm{J} / 29.25 \times 10^{-3}\ \mathrm{mol} = 56821.02\ \mathrm{J\ mol^{-1}}$

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