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Chapter 10: Problem 100

A \(0.250-\mathrm{g}\) chunk of sodium metal is cautiously dropped into a mixture of \(50.0 \mathrm{~g}\) water and \(50.0 \mathrm{~g}\) ice, both at \(0{ }^{\circ} \mathrm{C}\). The reaction is \(2 \mathrm{Na}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{NaOH}(a q)+\mathrm{H}_{2}(g) \quad \Delta H=-368 \mathrm{~kJ}\) Will the ice melt? Assuming the final mixture has a specific heat capacity of \(4.18 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\), calculate the final temperature. The enthalpy of fusion for ice is \(6.02 \mathrm{~kJ} / \mathrm{mol}\).

Short Answer

Expert verified
The ice will not melt completely as the heat released by the reaction of sodium metal with water is not enough to melt all the ice. The final temperature of the mixture will be 4.43掳C.

Step by step solution

01

Calculate the moles of sodium metal

Given that the mass of sodium metal is 0.250 g, we can find the moles of sodium metal by dividing the mass by its molar mass (23 g/mol): \[moles\:of\:Na = \frac{0.250\:g}{23\:g/mol} = 0.0109\:mol\]
02

Calculate the heat released by the reaction

Now, we can calculate the heat released by the reaction using the moles of sodium metal and the given reaction enthalpy: \[q = (moles\:of\:Na) \times (\frac{1\:mol\:H_2}{2\:mol\:Na}) \times \Delta H\] \[q = (0.0109\:mol) \times \frac{1}{2} \times (-368\:kJ) = -1.859\: kJ\]
03

Calculate the heat required to melt the ice

The heat required to melt the ice can be calculated using the enthalpy of fusion and the moles of water present in the ice: \[ Heat\:required = (moles\:of\:H_2O)\times \Delta H_{fusion}\] Since 1 mol of water has a mass of 18 g, we have: \[moles\:of\:H_2O = \frac{50.0\:g}{18\:g/mol} = 2.778\:mol\] \[Heat\:required = (2.778\:mol) \times (6.02 \: kJ/mol) = 16.716\:kJ\]
04

Determine if the ice will melt

Since the heat released by the reaction (-1.859 kJ) is less than the heat required to melt the ice (16.716 kJ), the ice does not melt completely.
05

Calculate the final temperature of the mixture

Now we have to calculate the final temperature of the mixture. We know that the specific heat capacity of the final mixture is 4.18 J/g掳C. Since the heat released from the reaction will be absorbed by the ice and water mixture, we can set up the following equation: \[q_{absorbed} = (mass_{ice} + mass_{water})\times c_{mixture}\times \Delta T\] \[\Delta T = \frac{-q_{reaction}}{(mass_{ice} + mass_{water})\times c_{mixture}} \] \[ \Delta T = \frac{1.859 \times 10^3\:J}{(50.0\:g + 50.0\:g) \times 4.18 \: J/g\cdot{ }^{\circ}C} = 4.43^{\circ}\:C\] Since the initial temperature is 0掳C, the final temperature is: \[ T_{final} = T_{initial} + \Delta T = 0^{\circ}C + 4.43^{\circ}C = 4.43{ }^{\circ}\mathrm{C}\] Thus, the answer is that the ice will not melt, and the final temperature of the mixture will be 4.43掳C.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Enthalpy of Reaction
Enthalpy of reaction is the heat change during a chemical reaction. In our exercise, the reaction between sodium metal and water releases energy. This is indicated by a negative enthalpy change \( \Delta H = -368 \, \mathrm{kJ} \). The negative sign means the reaction is exothermic, releasing heat into the surroundings.

To calculate the total heat released, we first find the moles of sodium used in the reaction. With \( 0.250 \, \mathrm{g} \) of sodium and its molar mass \( 23 \, \mathrm{g/mol} \), we determine the moles as \( 0.0109 \, \mathrm{mol} \). Multiplying these moles by the enthalpy change per mole for the balanced reaction yields the heat released: \( -1.859 \, \mathrm{kJ} \). This value tells us how much energy is released when the reaction occurs.
Enthalpy of Fusion
Enthalpy of fusion is the energy required to change a solid into a liquid at its melting point. For ice, this is \( 6.02 \, \mathrm{kJ/mol} \). It represents the energy needed to melt 1 mole of ice without changing the temperature.

The problem involves calculating whether enough heat is available to melt the ice. With ice weighing \( 50.0 \, \mathrm{g} \), we find the moles: \( \frac{50.0 \, \mathrm{g}}{18 \, \mathrm{g/mol}} = 2.778 \, \mathrm{mol} \). The energy required to melt this amount of ice is \( 16.716 \, \mathrm{kJ} \).

Since the reaction releases less energy than needed to melt the ice, not all the ice will turn into water.
Specific Heat Capacity
Specific heat capacity is the amount of heat energy required to raise the temperature of 1 gram of a substance by 1掳C. The specific heat capacity of water in our scenario is \( 4.18 \, \mathrm{J/g掳C} \), a standard value for water, which also applies to the ice-water mixture.

In our calculation, the combined mass of the ice and water is \( 100.0 \, \mathrm{g} \). Using the heat absorbed equation \( q = mc\Delta T \), where \( m \) is mass and \( c \) is specific heat, we solve for \( \Delta T \) to find the temperature change caused by the heat released from the reaction. This change is \( 4.43掳C \), which adds to the initial temperature to yield a final temperature.
Heat Transfer in Reactions
Heat transfer in reactions involves the movement of thermal energy from the reaction to the surrounding environment. Here, the exothermic reaction between sodium and water releases heat that is absorbed by the ice-water mixture, potentially causing a temperature rise.

In this exercise, the total energy released by the reaction was insufficient to melt all the ice. Instead, the heat transfer raises the temperature of the mixture. We calculate the change in temperature \( \Delta T \) by dividing the heat released by the total mass and specific heat capacity of the mixture.

Ultimately, this illustrates how energy conservation and transfer play crucial roles in determining the outcomes of physical and chemical processes, such as changes in temperature or phase.

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Most popular questions from this chapter

The \(\mathrm{CsCl}\) structure is a simple cubic array of chloride ions with a cesium ion at the center of each cubic array (see Exercise 67 ). Given that the density of cesium chloride is \(3.97 \mathrm{~g} / \mathrm{cm}^{3}\), and assuming that the chloride and cesium ions touch along the body diagonal of the cubic unit cell, calculate the distance between the centers of adjacent \(\mathrm{Cs}^{+}\) and \(\mathrm{Cl}^{-}\) ions in the solid. Compare this value with the expected distance based on the sizes of the ions. The ionic radius of \(\mathrm{Cs}^{+}\) is \(169 \mathrm{pm}\), and the ionic radius of \(\mathrm{Cl}^{-}\) is \(181 \mathrm{pm}\).

Atoms are assumed to touch in closest packed structures, yet every closest packed unit cell contains a significant amount of empty space. Why?

Dry nitrogen gas is bubbled through liquid benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) at \(20.0^{\circ} \mathrm{C}\). From \(100.0 \mathrm{~L}\) of the gaseous mixture of nitrogen and benzene, \(24.7 \mathrm{~g}\) benzene is condensed by passing the mixture through a trap at a temperature where nitrogen is gaseous and the vapor pressure of benzene is negligible. What is the vapor pressure of benzene at \(20.0{ }^{\circ} \mathrm{C}\) ?

The molar heat of fusion of sodium metal is \(2.60 \mathrm{~kJ} / \mathrm{mol}\), whereas its heat of vaporization is \(97.0 \mathrm{~kJ} / \mathrm{mol}\). a. Why is the heat of vaporization so much larger than the heat of fusion? b. What quantity of heat would be needed to melt \(1.00 \mathrm{~g}\) sodium at its normal melting point? c. What quantity of heat would be needed to vaporize \(1.00 \mathrm{~g}\) sodium at its normal boiling point? d. What quantity of heat would be evolved if \(1.00 \mathrm{~g}\) sodium vapor condensed at its normal boiling point?

The structure of manganese fluoride can be described as a simple cubic array of manganese ions with fluoride ions at the center of each edge of the cubic unit cell. What is the charge of the manganese ions in this compound?
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