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Chapter 18: Problem 29

Consider the decomposition of calcium carbonate. $$ \mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$$ Calculate the pressure in atm of \(\mathrm{CO}_{2}\) in an equilibrium process (a) at \(25^{\circ} \mathrm{C}\) and (b) at \(800^{\circ} \mathrm{C}\). Assume that \(\Delta H^{\circ}=177.8 \mathrm{~kJ} / \mathrm{mol}\) and \(\Delta S^{\circ}=160.5 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\) for the temperature range.

Short Answer

Expert verified
The pressure of CO2 in an equilibrium process will be equal to \(K_1\) at \(25^{\circ}C\) and \(K_2\) at \(800^{\circ}C\). The exact values for \(K_1\) and \(K_2\) will depend on the specific numerical calculations in the steps outlined above. Use these steps to find the specific values.

Step by step solution

01

Transform the temperatures

First, the given temperatures should be converted from Celsius to Kelvin, because thermodynamic calculations deal with Kelvin scale. The conversion from Celsius to Kelvin is done with the formula \(K = ^{\circ}C + 273.15\). Therefore, the temperatures in Kelvin are \(T_1 = 25 + 273.15 = 298.15 K\) and \(T_2 = 800 + 273.15 = 1073.15 K\).
02

Convert \(\Delta H\) and \(\Delta S\) into suitable units

The given \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) need to be in the same units before they can be used in the Van't Hoff equation. Given that \(\Delta H^{\circ} = 177.8 kJ/mol\), it can be converted to \(J/mol\) by multiplying by \(1000\), yielding \(\Delta H^{\circ} = 177800 J/mol\). The given \(\Delta S^{\circ}\) is already in the appropriate units \(J/(K \cdot mol)\) and doesn't need to be converted.
03

Applying Van't Hoff equation

Now, calculate the value of equilibrium constant \(K_1\) at \(T_1 = 298.15 K\) by applying the Van't Hoff equation, which in logarithmic form is \(\ln K_1 = -\Delta H^{\circ}/(RT_1) + \Delta S^{\circ}/R\). You can use a standard molar gas constant \(R = 8.314 J/(K \cdot mol)\). After doing the calculation, take exponent on both sides to get \(K_1\).
04

Calculate the pressure at 25C

As the only gaseous component is CO2, at \(T_1\), the pressure of CO2 can be considered as the equilibrium constant, because the pressure units of a gas can be expressed in terms of atmospheres, bar, or torr etc. Therefore, you can take \(P = K_1\) at \(T_1\).
05

Calculate the value of equilibrium constant \(K_2\) at \(T_2 = 1073.15 K\)

Once again, apply the Van't Hoff equation to find the equilibrium constant at the second temperature. However, this time the equation will look like: \(\ln K_2 = \ln K_1 - \Delta H^{\circ}/R \cdot (1/T_2 - 1/T_1)\). Calculate the logarithm value of \(K_2\) and then take exponent to get the value of \(K_2\), which at this temperature will represent the pressure of CO2.
06

Calculate the pressure at 800C

Similar to step 4, the equilibrium constant at 800C is equal to the pressure of CO2. Since the only gaseous product is CO2, the total pressure in the system at equilibrium is equal to the partial pressure of CO2 namely \(P = K_2\) at \(T_2\).

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Most popular questions from this chapter

Predict whether the entropy change is positive or negative for each of these reactions: (a) \(\mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g)\) (b) \(\mathrm{O}(g)+\mathrm{O}(g) \longrightarrow \mathrm{O}_{2}(g)\) (c) \(\mathrm{NH}_{4} \mathrm{NO}_{3}(s) \longrightarrow \mathrm{N}_{2} \mathrm{O}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) (d) \(2 \mathrm{H}_{2} \mathrm{O}_{2}(l) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)\)

The standard enthalpy of formation and the standard entropy of gaseous benzene are \(82.93 \mathrm{~kJ} / \mathrm{mol}\) and \(269.2 \mathrm{~J} / \mathrm{K} \cdot\) mol, respectively. Calculate \(\Delta H^{\circ}, \Delta S^{\circ}\) and \(\Delta G^{\circ}\) for the process at \(25^{\circ} \mathrm{C}\). Comment on your answers. $$ \mathrm{C}_{6} \mathrm{H}_{6}(l) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(g) $$

For reactions carried out under standard-state conditions, Equation (18.10) takes the form \(\Delta G^{\circ}=\Delta H^{\circ}-\) \(T \Delta S^{\circ} .\) (a) Assuming \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are independent of temperature, derive the equation $$\ln \frac{K_{2}}{K_{1}}=\frac{\Delta H^{\circ}}{R}\left(\frac{T_{2}-T_{1}}{T_{1} T_{2}}\right)$$ where \(K_{1}\) and \(K_{2}\) are the equilibrium constants at \(T_{1}\) and \(T_{2}\), respectively. (b) Given that at \(25^{\circ} \mathrm{C} K_{\mathrm{c}}\) is \(4.63 \times 10^{-3}\) for the reaction $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) \quad \Delta H^{\circ}=58.0 \mathrm{~kJ} / \mathrm{mol} $$ calculate the equilibrium constant at \(65^{\circ} \mathrm{C}\)

(a) Over the years there have been numerous claims about "perpetual motion machines," machines that will produce useful work with no input of energy. Explain why the first law of thermodynamics prohibits the possibility of such a machine existing. (b) Another kind of machine, sometimes called a “perpetual motion of the second kind," operates as follows. Suppose an ocean liner sails by scooping up water from the ocean and then extracting heat from the water, converting the heat to electric power to run the ship, and dumping the water back into the ocean. This process does not violate the first law of thermodynamics, for no energy is created-energy from the ocean is just converted to electrical energy. Show that the second law of thermodynamics prohibits the existence of such a machine.

For each pair of substances listed here, choose the one having the larger standard entropy value at \(25^{\circ} \mathrm{C}\). The same molar amount is used in the comparison. Explain the basis for your choice. (a) \(\operatorname{Li}(s)\) or \(\operatorname{Li}(l)\) (b) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)\) or \(\mathrm{CH}_{3} \mathrm{OCH}_{3}(l)\) (c) \(\operatorname{Ar}(g)\) or \(\operatorname{Xe}(g)\) (d) \(\mathrm{CO}(g)\) or \(\mathrm{CO}_{2}(g)\) (e) \(\mathrm{O}_{2}(g)\) or \(\mathrm{O}_{3}(g)\) (f) \(\mathrm{NO}_{2}(g)\) or \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\)
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