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Chapter 5: Problem 53

When a 9.55-g sample of solid sodium hydroxide dissolves in \(100.0 \mathrm{~g}\) of water in a coffee-cup calorimeter (Figure 5.17), the temperature rises from \(23.6^{\circ} \mathrm{C}\) to \(47.4^{\circ} \mathrm{C}\). Calculate \(\Delta H\) (in \(\mathrm{kJ} / \mathrm{mol} \mathrm{NaOH}\) ) for the solution process $$ \mathrm{NaOH}(s) \stackrel{-\cdots}{\mathrm{Na}}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ Assume that the specific heat of the solution is the same as that of pure water.

Short Answer

Expert verified
The enthalpy change for the solution process of NaOH is approximately \(41.7\,\mathrm{kJ/mol}\).

Step by step solution

01

1. Calculate the amount of heat exchange

First, we need to determine the heat exchanged between the sodium hydroxide and water using the specific heat capacity of water, the mass of water in the calorimeter, and the change in temperature. The heat exchange (q) can be calculated using the formula: \(q = mc\Delta T\) Where: - q is the heat exchange - m is the mass of the substance (here, the mass of water) - c is the specific heat capacity (4.18 J/g掳C for water) - 螖T is the change in temperature
02

2. Determine the change in temperature

To calculate 螖T (the change in temperature), subtract the initial temperature from the final temperature: \(螖T = T_{final} - T_{initial}\) Plug in the given values: \(螖T = 47.4\,^{\circ}\mathrm{C} - 23.6^{\circ} \mathrm{C} = 23.8\,^{\circ}\mathrm{C}\)
03

3. Calculate the heat exchange

Now that 螖T is known, we can calculate the heat exchange using the formula from step 1. We know the mass of water is 100 g and the specific heat capacity is 4.18 J/g掳C: \(q = (100\,\mathrm{g})(4.18\,\mathrm{J}/ \mathrm{g}\,^{\circ}\mathrm{C})(23.8\,^{\circ}\mathrm{C})\) \(q = 9954.4\,\mathrm{J}\) or \(9.954\,\mathrm{kJ}\)
04

4. Find the number of moles of NaOH

We now need to determine the number of moles of NaOH in the given 9.55 g sample. To do that, we can use the molecular weight of NaOH (鈮�40 g/mol) and the following formula: \(moles\,of\,NaOH = \frac{mass\,of\,NaOH}{molecular\,weight\,of\,NaOH}\) \(moles\,of\,NaOH = \frac{9.55\,\mathrm{g}}{40\,\mathrm{g/mol}}\) \(moles\,of\,NaOH = 0.23875\,\mathrm{moles}\)
05

5. Calculate the enthalpy change per mole of NaOH

Finally, to find the enthalpy change per mole of NaOH, divide the heat exchange (q) from step 3 by the moles of NaOH from step 4: \(\Delta H = \frac{q}{moles\,of\,NaOH}\) \(\Delta H = \frac{9.954\,\mathrm{kJ}}{0.23875\,\mathrm{moles}}\) \(\Delta H \approx 41.7\,\mathrm{kJ/mol}\) The enthalpy change for the solution process of NaOH is approximately 41.7 kJ/mol.

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Most popular questions from this chapter

Limestone stalactites and stalagmites are formed in caves by the following reaction: \(\mathrm{Ca}^{2+}(a q)+2 \mathrm{HCO}_{3}^{-}(a q) \longrightarrow\) \(\mathrm{CaCO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(I)\) If \(1 \mathrm{~mol}\) of \(\mathrm{CaCO}_{3}\) forms at \(298 \mathrm{~K}\) under \(1 \mathrm{~atm}\) pressure, the reaction performs \(2.47 \mathrm{~kJ}\) of \(P-V\) work, pushing back the atmosphere as the gaseous \(\mathrm{CO}_{2}\) forms. At the same time, \(38.95 \mathrm{~kJ}\) of heat is absorbed from the environment. What are the values of \(\Delta H\) and of \(\Delta E\) for this reaction?

Consider the combustion of liquid methanol, \(\mathrm{CH}_{3} \mathrm{OH}(I)\) : \(\mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\) \(\Delta H=-726.5 \mathrm{~kJ}\) (a) What is the enthalpy change for the reverse reaction? (b) Balance the forward reaction with whole-number coefficients. What is \(\Delta H\) for the reaction represented by this equation? (c) Which is more likely to be thermodynamically favored, the forward reaction or the reverse reaction? (d) If the reaction were written to produce \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) instead of \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\), would you expect the magni- tude of \(\Delta H\) to increase, decrease, or stay the same? Explain.

Suppose you toss a tennis ball upward. (a) Does the kinetic energy of the ball increase or decrease as it moves higher? (b) What happens to the potential energy of the ball as it moves higher? (c) If the same amount of energy were imparted to a ball the same size as a tennis ball, but of twice the mass, how high would it go in comparison to the tennis ball? Explain your answers.

Consider the combustion of a single molecule of \(\mathrm{CH}_{4}(g)\) forming \(\mathrm{H}_{2} \mathrm{O}(l)\) as a product. (a) How much energy, in J. is produced during this reaction? (b) A typical X-ray photon has an energy of \(8 \mathrm{keV}\). How does the energy of combustion compare to the energy of the X-ray photon?

Consider the twodiagramsbelow. (a) Based on \((t)\), write an equation showing how \(\Delta H_{\mathrm{A}}\) is related to \(\Delta H_{\mathrm{B}}\) and \(\Delta H_{\mathrm{C}}\). How do both diagram (i) and your equation relate to the fact that enthalpy is a state function? (b) Based on (ii), write an equation relating \(\Delta H_{Z}\) to the other enthalpy changes in the diagram. (c) How do these diagrams relate to Hess's law? [Section 5.6]
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