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Chapter 2: Problem 36

A pellet of Zn of mass \(31.2 \mathrm{g}\) is dropped into a flask containing dilute \(\mathrm{H}_{2} \mathrm{SO}_{4}\) at a pressure of \(P=1.00\) bar and a temperature of \(T=300 .\) K. What is the reaction that occurs? Calculate \(w\) for the process.

Short Answer

Expert verified
The reaction between zinc (Zn) and sulfuric acid (H鈧係O鈧�) is given by the balanced chemical equation: \(Zn(s) + H_{2}SO_{4}(aq) \rightarrow ZnSO_{4}(aq) + H_{2}(g)\). Given the mass of Zn as 31.2 g, the number of moles is found to be 0.477 moles of both Zn and H鈧� gas. The energy of the system with pressure 100,000 Pa and temperature 300 K is calculated as 1193.42 J using Ideal Gas Law. The work (w) for the process, determined by calculating the change in volume, is 0 J.

Step by step solution

01

Write the Balanced Chemical Equation

The reaction that occurs when Zn (zinc) comes in contact with H鈧係O鈧� (sulfuric acid) is: \[Zn(s) + H_{2}SO_{4}(aq) \rightarrow ZnSO_{4}(aq) + H_{2}(g)\] Now, let's calculate the number of moles of Zn and the energy for the given pressure and temperature.
02

Determine Moles and Energy

Given that the mass of Zn is 31.2 g. To calculate the number of moles (n), use the molar mass of Zn (65.38 g/mol): \[n = \frac{mass}{molar \thinspace mass} = \frac{31.2 \thinspace g}{65.38 \thinspace g/mol} = 0.477 \thinspace moles\] Using the balanced equation, we obtain 0.477 moles of H鈧� gas. Pressure (P) = 1.00 bar = 100,000 Pa and Temperature (T) = 300 K. We need to convert the pressure and temperature into energy using the gas constant (R = 8.314 J/mol路K): \[\Delta U = nRT = (0.477 \thinspace moles)(8.314 J/mol \cdot K)(300 \thinspace K) = 1193.42 \thinspace J\]
03

Apply Ideal Gas Law to Calculate Work

Next, we will use the ideal gas law (PV=nRT) to calculate the change in volume and work done by the system. First, we need to calculate the initial and final volume of H鈧� gas in the flask. Initial Volume (V鈧�): \[V_{1} = \frac{nRT}{P} = \frac{(0.477 \thinspace moles)(8.314 J/mol \cdot K)(300 \thinspace K)}{100,000 \thinspace Pa} = 0.01190 \thinspace m^3\] Since all of the Zn reacts with H鈧係O鈧� and exists only as ZnSO鈧� and H鈧� gas with the same pressure and temperature, the final volume of the system (V鈧�) remains the same as V鈧�. Now, we can calculate the work done by the system: \[\Delta V = V_{2} - V_{1} = 0\] \[w = -P\Delta V = -(100,000 \thinspace Pa)(0 \thinspace m^3) = 0 \thinspace J\] The work (w) for the process is 0 J.

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Key Concepts

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

Balanced Chemical Equation
In the context of chemical reactions, a balanced chemical equation ensures that the number of each type of atom on the reactant side is equal to the number on the product side. This stems from the law of conservation of mass, which indicates that matter cannot be created or destroyed.

For the reaction between zinc (Zn) and sulfuric acid (H鈧係O鈧�), the challenge is to write a balanced equation representing each substance involved in the process.
Here, the balanced chemical equation is given by:
\[Zn(s) + H_{2}SO_{4}(aq) \rightarrow ZnSO_{4}(aq) + H_{2}(g)\]

This equation shows that one mole of solid zinc reacts with one mole of aqueous sulfuric acid to produce one mole of zinc sulfate and one mole of hydrogen gas.
  • Reactants: Zn and H鈧係O鈧�
  • Products: ZnSO鈧� and H鈧�
When balancing chemical equations, it's important to adjust the coefficients 鈥� the numbers in front of each molecule 鈥� and not change the subscripts of the formulas, as this would change the identity of the substances.
Ideal Gas Law
The Ideal Gas Law is a fundamental equation that relates the pressure, volume, temperature, and number of moles of a gas. It's expressed as \(PV = nRT\), where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the universal gas constant (8.314 J/mol路K), and \(T\) is the temperature in Kelvin.

In the zinc and sulfuric acid reaction, hydrogen gas is produced. Knowing the number of moles ( \(n = 0.477 \, ext{moles}\)) of hydrogen gas allows us to use the Ideal Gas Law to find the volume it occupies under certain conditions.
Using the equation:
\[ V = \frac{nRT}{P} \]

Given that the pressure \(P = 100,000 \, ext{Pa}\) and the temperature \(T = 300 \, ext{K}\), we can calculate the initial volume provided all these parameters are constant during the reaction.
  • This helps us understand gas behavior under the given conditions.
  • Remember, it's idealized and not all gases behave perfectly under all conditions.
This simplification allows for easier calculations when predicting the behavior of gas in chemical reactions.
Moles and Energy Calculation
Calculating moles is crucial for understanding chemical reactions and determining the amounts of reactants and products involved. The number of moles \(n\) of a substance is calculated using its mass divided by its molar mass.

For zinc in our reaction:
\[n = \frac{31.2 \, ext{g}}{65.38 \, ext{g/mol}} = 0.477 \, ext{moles} \]

This tells us how much zinc and ultimately how much hydrogen gas will be produced.
Subsequently, the energy change \((\Delta U)\) can be calculated using \(\Delta U = nRT\), where all variables are known.
This energy calculation reflects the internal energy change of the system.

In physical chemistry, such calculations are fundamental for understanding reaction dynamics and energy flow.
  • It helps in determining how much energy is needed or released.
  • This can influence reaction conditions or the need for catalysts.
Keep in mind, these calculations assume ideal conditions, and real-world deviations might occur.

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Most popular questions from this chapter

DNA can be modeled as an elastic rod that can be twisted or bent. Suppose a DNA molecule of length \(L\) is bent such that it lies on the arc of a circle of radius \(R_{c}\). The reversible work involved in bending DNA without twisting is \(w_{b e n d}=\frac{B L}{2 R_{c}^{2}}\) where \(B\) is the bending force constant. The DNA in a nucleosome particle is about \(680 . \AA\) in length. Nucleosomal DNA is bent around a protein complex called the histone octamer into a circle of radius 55 A. Calculate the reversible work involved in bending the DNA around the histone octamer if the force constant \(B=2.00 \times 10^{-28} \mathrm{Jm}\)

At \(298 \mathrm{K}\) and 1 bar pressure, the density of water is \(0.9970 \mathrm{g} \mathrm{cm}^{-3},\) and \(C_{P, m}=75.3 \mathrm{JK}^{-1} \mathrm{mol}^{-1} .\) The change in volume with temperature is given by \(\Delta V=V_{\text {initial}} \beta \Delta T\) where \(\beta,\) the coefficient of thermal expansion, is \(2.07 \times 10^{-4} \mathrm{K}^{-1}\). If the temperature of \(325 \mathrm{g}\) of water is increased by \(25.5 \mathrm{K}\), calculate \(w, q, \Delta H,\) and \(\Delta U\)

A vessel containing 1.50 mol of an ideal gas with \(P_{i}=1.00\) bar and \(C_{P, m}=5 R / 2\) is in thermal contact with a water bath. Treat the vessel, gas, and water bath as being in thermal equilibrium, initially at \(298 \mathrm{K},\) and as separated by adiabatic walls from the rest of the universe. The vessel, gas, and water bath have an average heat capacity of \(C_{P}=2450 . \mathrm{JK}^{-1} .\) The gas is compressed reversibly to \(P_{f}=20.5\) bar. What is the temperature of the system after thermal equilibrium has been established?

A \(2.50 \mathrm{mol}\) sample of an ideal gas for which \(C_{V, m}=3 R / 2\) undergoes the following two-step process: (1) From an initial state of the gas described by \(T=13.1^{\circ} \mathrm{C}\) and \(P=1.75 \times 10^{5} \mathrm{Pa}\), the gas undergoes an isothermal expansion against a constant external pressure of \(3.75 \times 10^{4} \mathrm{Pa}\) until the volume has doubled. (2) Subsequently, the gas is cooled at constant volume. The temperature falls to \(-23.6^{\circ} \mathrm{C}\). Calculate \(q, w, \Delta U,\) and \(\Delta H\) for each step and for the overall process.

A 3.50 mole sample of \(\mathrm{N}_{2}\) in a state defined by \(T_{i}=\) 250\. \(\mathrm{K}\) and \(V_{i}=3.25 \mathrm{L}\) undergoes an isothermal reversible expansion until \(V_{f}=35.5 \mathrm{L}\) Calculate \(w,\) assuming (a) that the gas is described by the ideal gas law, and (b) that the gas is described by the van der Waals equation of state. What is the percent error in using the ideal gas law instead of the van der Waals equation? The van der Waals parameters for \(\mathrm{N}_{2}\) are listed in Table 7.4
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