\(1.00\) litre sample of a mixture of \(\mathrm{CH}_{4(\mathrm{~g})}\) and \(\mathrm{O}_{2(\mathrm{~g})}\) measured at \(25^{\circ} \mathrm{C}\) and 740 torr was allowed to react at constant pressure in a calorimeter which together with its contents had a heat capacity of \(1260 \mathrm{cal} / \mathrm{K}\). The complete combustion of the methane to \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) caused a temperature rise in the calorimeter of \(0.667 \mathrm{~K}\). What was the mole per cent of \(\mathrm{CH}_{4}\) in the original mixture? \(\Delta H_{\text {comb }}^{\circ}\left(\mathrm{CH}_{4}\right)=-215 \mathrm{k}\) cal \(\mathrm{mol}^{-1}\).

Short Answer

Expert verified
The mole percent of CH4 in the original mixture can be found by dividing the moles of combusted CH4, as determined from the heat released, by the initial moles of the gas mixture, and then multiplying by 100%. After performing the necessary conversions and calculations, this will yield the mole percent of CH4.

Step by step solution

01

Determine the Partial Pressure of CH4

The partial pressure of methane (CH4) can be found by using Dalton's Law of Partial Pressures. Dalton's Law states that the total pressure of a gas mixture is the sum of the partial pressures of each individual gas in the mixture. If we assume that the entire pressure change is due to the combustion of CH4, then initially the partial pressure of CH4 is the same as the total pressure, which is 740 torr.
02

Convert Pressure to Atmospheres

Since the combustion enthalpy is often given in calories per mole at standard conditions (1 atm pressure), we need to convert the pressure of methane from torr to atmospheres. Use the conversion factor that 1 atm = 760 torr to make this conversion.
03

Calculate Moles of CH4

Use the ideal gas law, which states that PV = nRT, to calculate the number of moles of CH4. You'll need to use the pressure in atmospheres, the volume in liters, the gas constant R (0.0821 L atm K^-1 mol^-1), and the temperature in Kelvin. Note that the temperature in Celsius must be converted to Kelvin by adding 273.15, so T = 25 + 273.15.
04

Calculate the Heat Produced in the Reaction

The heat produced is a result of the combustion of methane and can be calculated by the formula q = mcΔT, where m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. In this case, the calorimeter provides us with the constant volume heat capacity (C), and the change in temperature is given. So, the heat released (q) is CΔT.
05

Calculate Energy Change Per Mole of CH4

Now that you have the total heat produced, use the combustion enthalpy of CH4 to find out how many moles of CH4 were combusted. The enthalpy change of combustion tells us the amount of heat released when one mole of a substance is combusted. The formula is ΔH = q/n, where q is the heat released and n is the moles of CH4 burnt.
06

Calculate Mole Percentage

To find the mole percentage of CH4 in the original mixture, divide the moles of CH4 (found from the heat calculation) by the total moles of the gas mixture (from the ideal gas law) and multiply by 100%.

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Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is a fundamental principle in chemistry that relates the pressure (P), volume (V), temperature (T), and number of moles (n) of an ideal gas. Expressed by the equation PV = nRT, where R is the gas constant, this law allows us to calculate any one of these variables if the others are known.

For the problem at hand, we used the Ideal Gas Law to determine the amount of methane in moles. Knowing the volume of the sample, the temperature in Kelvin, and the pressure in atmospheres, we could solve for n, the number of moles of methane present before combustion. It's important to convert all measurements to their standard units: liters for volume, Kelvin for temperature, and atmospheres for pressure, to ensure the correct application of the gas constant, R.

We must also remember that real gases may not behave exactly like ideal gases, but the Ideal Gas Law provides a good approximation for many conditions, particularly those close to standard temperature and pressure.
Calorimetry
Calorimetry is the measurement of heat flow in a chemical reaction, physical change, or heat capacity. In our exercise, we are using calorimetry to find out the amount of heat produced by the combustion of methane.The calorimeter had a known heat capacity, and by observing the change in temperature, we could calculate the heat released using the formula q = mcΔT. Because we were given a constant pressure calorimeter, the heat capacity used in the calculation was at constant pressure. The value CΔT gave us the total heat q released by the reaction. This step is critical to understanding how much energy is produced by the combustion of a known amount of methane. Understanding calorimetry is essential in fields that examine energy changes in chemical processes, such as thermodynamics.
Dalton's Law of Partial Pressures
Dalton's Law of Partial Pressures is a gas law that states that in a mixture of non-reacting gases, the total pressure exerted is equal to the sum of the partial pressures of the individual gases. This law implies that gases in a mixture each exert pressure independently of each other.In this exercise, we assumed that all the pressure was due to methane before the reaction. Since methane and oxygen were the only gases present and they react completely, the initial partial pressure of methane can be determined from the total pressure. This is an essential step because it sets the stage for using the Ideal Gas Law to calculate the moles of methane in the reaction. Dalton's Law is an important concept in calculating the composition of gas mixtures and understanding the behavior of gases in various conditions.
Enthalpy of Combustion
Enthalpy of combustion is a chemical property that signifies the heat released when one mole of a substance burns completely in the presence of oxygen. It is usually expressed in energy units per mole, such as kilocalories per mole (kcal/mol). The negative sign indicates the release of heat, making the process exothermic.In the problem provided, the combustion enthalpy of methane is given as -215 kcal/mol. This value helped us determine the amount of heat released during the reaction. By dividing the total heat released by this enthalpy value, we can find the number of moles of methane that combusted. This step is instrumental to finding the mole percent of methane in the original mixture, which ties back into our understanding of calorimetry and the Ideal Gas Law. Enthalpy changes, such as the enthalpy of combustion, are vital for quantifying energy changes in reactions and for gauging the efficiency of fuel sources.

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Most popular questions from this chapter

The integral enthalpy of solution in \(\mathrm{kJ}\) of one mole of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) dissolved in \(n\) mole of water is given by: $$ \Delta H_{s}=\frac{75.6 \times n}{n+1.8} $$ Calculate \(\Delta H\) for the following process: (a) 1 mole of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) dissolved in \(2 \mathrm{~mole}\) of \(\mathrm{H}_{2} \mathrm{O}\). (b) 1 mole of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) dissolved in 7 mole of \(\mathrm{H}_{2} \mathrm{O}\). (c) 1 mole of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) dissolved in 5 mole of \(\mathrm{H}_{2} \mathrm{O}\). (d) solution (a) dissolved in 5 mole of \(\mathrm{H}_{2} \mathrm{O}\). (e) 1 mole of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) dissolved in excess of \(\mathrm{H}_{2} \mathrm{O}\).

The standard enthalpy of formation of \(\mathrm{FeO}\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) is \(-65 \mathrm{kcal}\) \(\mathrm{mol}^{-1}\) and \(-197\) kcal mol \(^{-1}\) respectively. A mixture of two oxides contains \(\mathrm{FeO}\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) in the mole ratio \(2: 1 .\) If by oxidation, it is changed into a \(1: 2\) mole ratio mixture, how much of thermal energy will be released per mol of initial mixture.

The heat of solution of \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) in water was determined by measuring the amount of electrical work needed to compensate for the cooling which would otherwise óccur when the salt dissolves. After the \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) was added to the water, electrical energy was provided by passage of current through a resistance coil until the temperature of the solution reached the value it had prior to the addition of salt. In a typical experiment, \(4.4 \mathrm{~g}\) of \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) was added to \(200 \mathrm{~g}\) water. A current of \(0.75\) ampere was provided through the heater coil, and the voltage across the terminals was \(6.0 \mathrm{~V}\). The current was applied for \(5.2\) minute. Calculate \(\Delta H\) for the solution of \(1.0\) mole \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) in enough water to give same concentration as was attained in the above experiment.

The commercial production of water gas utilizes the reaction under standard conditions: \(\mathrm{C}+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{H}_{2}+\mathrm{CO}\). The heat required for this endothermic reaction may be supplied by adding a limited amount of air and burning some carbon to \(\mathrm{CO}_{2}\). How many \(\mathrm{g}\) of carbon must be burnt to \(\mathrm{CO}_{2}\) to provide enough heat for the water gas conversion of \(100 \mathrm{~g}\) carbon? Neglect all heat losses to the environment. Also \(\Delta H_{\mathrm{f}}^{2}\) of \(\mathrm{CO}, \mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})}\) and \(\mathrm{CO}_{2}\) are \(-110.53\), \(-241.81\) and \(-393.51 \mathrm{~kJ} / \mathrm{mol}\) respectively.

\begin{aligned} &\text { The heat of dissociation of } \mathrm{H}_{2} \text { is } 435 \mathrm{~kJ} \mathrm{~mol}^{-1} \text { . If } \mathrm{C}_{(\mathrm{s})} \longrightarrow \mathrm{C}_{(\mathrm{g})} \text { ; }\\\ &\Delta H=720 \mathrm{~kJ} \mathrm{~mol}^{-1}, \text { calculate the bond energy per mol of the } \mathrm{C}-\mathrm{H} \text { bond }\\\ &\text { in } \mathrm{CH}_{4} \text { molecule. } \Delta H_{\mathrm{f}} \text { for } \mathrm{CH}_{4}=-75 \mathrm{~kJ} \mathrm{~mol}^{-1} \text { . } \end{aligned}

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