Problem 102 - Chemistry An Atoms First Approach

Question

A \(0.250-g\) 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 $$2 \mathrm{Na}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{NaOH}(a q)+\mathrm{H}_{2}(g) \quad \Delta H=-368 \mathrm{kJ}$$ Assuming no heat loss to the surroundings, 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 \(6.02 \mathrm{kJ} / \mathrm{mol}\).

Step-by-Step Solution

Answer

The heat released by the reaction (-4.00 kJ) is not enough to melt all the ice, as it requires 16.7 kJ to do so. Thus, some ice will melt, but not all. The final temperature of the mixture will be -1.09°C.

Step 1: Calculate moles of sodium

The first thing to do is to determine the number of moles of sodium in the 0.250g chunk. To do this, we need the molar mass of sodium, which is 22.99g/mol. The formula to calculate the moles of a substance is: moles = mass of substance / molar mass of substance moles of Na = \( \frac{0.250g}{22.99g/mol} \approx 0.0109\) moles

Step 2: Calculate the heat released by the reaction

Now we can calculate the heat released by the reaction. The enthalpy change, \(\Delta H\), tells us how much heat is released per mole of sodium, which is -368 kJ/mol. To find the total heat released, we can use the following formula: Q = moles of Na x \(\Delta H\) Q = (0.0109 moles) x (-368 kJ/mol) = -4.00 kJ This negative value indicates that heat is released in the reaction.

Step 3: Calculate the heat needed to melt the ice

Next, we need to calculate the heat required to melt the ice. We can use the enthalpy of fusion for ice, which is 6.02 kJ/mol, and the molar mass of water, which is 18.02 g/mol, in the following formula: Q = (mass of ice) x (\(\frac{1\ mol}{18.02\ g}\)) x (6.02 kJ/mol) Q = (50.0g) x (\(\frac{1\ mol}{18.02\ g}\)) x (6.02 kJ/mol) = 16.7 kJ

Step 4: Determine if the ice will melt

Now that we have calculated the heat released by the reaction (-4.00 kJ) and the heat needed to melt the ice (16.7 kJ), we can compare them. Since 4.00 kJ < 16.7 kJ, the heat released by the reaction is not enough to melt all the ice. Some ice will melt, but not all.

Step 5: Calculate the final temperature of the mixture

To find the final temperature of the mixture, we need to consider the heat absorbed by the remaining ice and water and compare it to the heat released by the reaction. First, let's calculate the mass of ice melted: Mass of melted ice = \(\frac{4.00\ kJ}{6.02\ kJ/mol}\) x \(18.02\ g/mol\) = 11.9 g Next, we can determine the mass of the remaining ice and water: Mass of remaining ice = 50.0 g - 11.9 g = 38.1 g Mass of water = 50.0 g + 11.9 g = 61.9 g Now, we can use the specific heat capacity of the mixture, 4.18 J/g°C, to calculate the final temperature: \(\Delta Q = (m_{ice} + m_{water})\times C_p\times \Delta T\) -4.00 kJ = (38.1 g + 61.9 g) x (4.18 J/g°C) x \(\Delta T\) Now, solve for \(\Delta T\): \(\Delta T =\frac{-4.00\ 10^3\ J}{(38.1\ g + 61.9\ g)(4.18\ J/g°C)} = -1.09\,°C\) Since the initial temperature of the mixture is 0°C: Final temperature = Initial temperature + \(\Delta T\) Final temperature = 0°C - 1.09°C = -1.09°C So, the final temperature of the mixture is -1.09°C.

Key Concepts

Thermochemistry

Thermochemistry is the branch of chemistry that deals with the energy changes associated with chemical reactions. In the context of the given exercise, understanding thermochemistry is essential to predict whether a reaction will release or absorb energy, and if so, how much. When sodium reacts with water, the process described is exothermic, meaning it releases energy; specifically, it releases heat.

An important quantity in thermochemistry is the enthalpy change \text(denoted as \( \textDelta H \)), which represents the heat absorbed or released by a chemical reaction at constant pressure. It's measured in kilojoules per mole (kJ/mol). A negative enthalpy change, as seen in the exercise (\( \textDelta H = -368 \text{ kJ/mol} \)), indicates that heat is released to the surroundings, and the reaction is exothermic. This release or absorption of heat can affect the surrounding materials - in this case, the ice and water - which sets the stage for calculating the outcome, such as the melting of ice.

Molar Mass

Molar mass is a fundamental concept in chemistry, representing the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It's invaluable for converting between the mass of a substance and the amount of substance in moles, which is imperative for stoichiometric calculations in chemical reactions.

In our exercise, we needed the molar mass of sodium to find out how many moles of sodium we had in a 0.250g chunk. With sodium's molar mass being 22.99 g/mol, we used the formula: \( \text{moles} = \frac{\text{mass of substance}}{\text{molar mass of substance}} \). Therefore, the molar mass acts as a bridge between the macroscopic scale of grams that we can measure and the microscopic scale of moles that we use for chemical equation balancing and reaction energetics.

Specific Heat Capacity

Specific heat capacity is a property that describes how much heat energy is required to raise the temperature of one gram of a substance by one degree Celsius (1°C). It’s a vital factor when considering energy changes in a substance due to heating or cooling. The unit for specific heat capacity is joules per gram per degree Celsius (J/g°C).

The significance of specific heat capacity is highlighted in the latter part of the problem where we need to calculate the final temperature of the water-ice mixture after the exothermic reaction. The mixture’s specific heat capacity (4.18 J/g°C for water) allows us to calculate the temperature change using the formula: \( \textDelta Q = (m_{\text{ice}} + m_{\text{water}}) \times C_p \times \textDelta T \).

Due to water's high specific heat capacity, it can absorb a significant amount of heat with minimal temperature change, making it an excellent buffer in this reaction. This also explains why all the ice doesn’t melt despite the exothermic nature of the sodium-water reaction.