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Assume that \(4.19 \times 10^{6} \mathrm{kJ}\) of energy is needed to heat a home. If this energy is derived from the combustion of methane \(\left(\mathrm{CH}_{4}\right),\) what volume of methane, measured at STP, must be burned? \(\left(\Delta H_{\text { combustion }}^{\circ} \text { for } \mathrm{CH}_{4}=-891 \mathrm{kJ} / \mathrm{mol}\right)\)

Short Answer

Expert verified
The volume of methane required to heat the home can be calculated using the given energy needed and the standard enthalpy change for the combustion of methane as follows: Volume of methane = \(\frac{4.19 \times 10^{6} kJ}{-891 kJ/mol}\) × \(22.4 L/mol\) By calculating the above expression, we find the required volume of methane that must be burned to provide the necessary energy to heat the home.

Step by step solution

01

Calculate the number of moles of methane needed

First, let's determine how many moles of methane are required to provide the given energy. We know that the energy needed is \(4.19 \times 10^{6} kJ\), and the standard enthalpy change for the combustion of methane is \(-891 kJ/mol\). To find the number of moles, we'll use the formula: Moles of methane = \(\frac{\text{Energy needed}}{\Delta H_{combustion}}\) Moles of methane = \(\frac{4.19 \times 10^{6} kJ}{-891 kJ/mol}\)
02

Calculate the volume of methane required at STP

Now, we need to convert the moles of methane into a volume, which we can do using the molar volume of a gas at STP. At STP (standard temperature and pressure), the molar volume of any gas is \(22.4 L/mol\). To find the volume of methane required, we'll use the formula: Volume of methane = Moles of methane × Molar volume at STP Volume of methane = \(\frac{4.19 \times 10^{6} kJ}{-891 kJ/mol}\) × \(22.4 L/mol\) Once you have calculated the volume of methane, you will have the required amount of methane that must be burned to heat the home with the given energy.

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Key Concepts

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

Molar Volume of Gas
The molar volume of a gas is a crucial concept in understanding how gases behave under specific conditions. Put simply, it is the volume that one mole of any gas occupies. At standard conditions, this value is constant. This means it is the same no matter the type of gas, provided these conditions remain unchanged.

At standard temperature and pressure (STP), the molar volume is exactly 22.4 liters per mole. This allows us to easily convert moles of gas into volume and vice versa.
  • This concept is particularly useful when dealing with chemical reactions where gases are involved.
  • It simplifies calculations, especially when predicting the volume of gas produced or consumed in a reaction.
Standard Temperature and Pressure (STP)
Standard temperature and pressure, or STP, is an important reference point in chemistry. It standardizes conditions to make measurements and calculations easier and more comparable. At STP, the temperature is set at 0 degrees Celsius, or 273.15 Kelvin, and the pressure is set at 1 atmosphere (atm).

These conditions mimic a baseline that scientists and engineers can rely on for accuracy and consistency. Under these standardized conditions, one mole of any ideal gas will occupy 22.4 liters, which is essential for converting between amounts of substance and volumes when working with gases.
  • The importance of STP is emphasized when calculating reactions in steps, ensuring that all participants agree on the conditions.
  • These standard conditions help in eliminating variable outcomes due to fluctuating environmental factors.
Moles of Methane
In chemistry, the concept of moles is central to quantifying substances involved in reactions. Methane is a simple hydrocarbon, often used as a fuel in combustion reactions. To convert energy needs into chemical amounts, we must calculate the number of moles of methane required for the reaction.

The formula to find moles when given energy is:\[\text{Moles of methane} = \frac{\text{Energy needed}}{\Delta H_{combustion}}\]This equation allows us to determine how much methane is needed. By using the known enthalpy of combustion (\(\Delta H_{combustion} = -891 \text{kJ/mol}\)), we find the required number of moles. This then helps in converting to volume at STP using the molar volume.
  • This process highlights the importance of understanding and converting units appropriately in chemistry.
  • It shows how energy requirements translate directly into chemical quantities.
Energy Conversion in Chemistry
Energy conversion is a fundamental aspect of chemical processes. It involves the transformation of energy from one form to another. In combustion, chemical energy stored in bonds is converted into heat energy, making this process central in heating and power generation.

The enthalpy of combustion represents the heat change when a mole of substance burns completely in oxygen. For methane, this is \(-891 \text{kJ/mol}\). This value guides calculations for determining how much energy can be harnessed from given amounts of fuel.
  • It is essential to understand energy conversion for tasks like designing efficient fuel utilization systems.
  • Knowing the energy output helps in planning and calculating energy input for homes or industries.
Thus, energy conversion principles allow us to match energy needs with the right chemical processes.

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Most popular questions from this chapter

Which of the following substances have an enthalpy of formation equal to zero? a. \(C l_{2}(g)\) b. \(\mathrm{H}_{2}(g)\) c. \(\mathrm{N}_{2}(l)\) d. \(\mathrm{Cl}(g)\)

The enthalpy of combustion of \(\mathrm{CH}_{4}(g)\) when \(\mathrm{H}_{2} \mathrm{O}(l)\) is formed is \(-891 \mathrm{kJ} / \mathrm{mol}\) and the enthalpy of combustion of \(\mathrm{CH}_{4}(g)\) when \(\mathrm{H}_{2} \mathrm{O}(g)\) is formed is \(-803 \mathrm{kJ} / \mathrm{mol}\) . Use these data and Hess's law to determine the enthalpy of vaporization for water.

Consider 2.00 moles of an ideal gas that are taken from state \(A\) \(\left(P_{A}=2.00 \mathrm{atm}, V_{A}=10.0 \mathrm{L}\right)\) to state \(B\left(P_{B}=1.00 \mathrm{atm}, V_{B}=\right.\) 30.0 \(\mathrm{L}\) ) by two different pathways: These pathways are summarized on the following graph of \(P\) versus \(V :\) Calculate the work (in units of \(\mathrm{J} )\) associated with the two path- ways. Is work a state function? Explain.

One of the components of polluted air is NO. It is formed in the high- temperature environment of internal combustion engines by the following reaction: $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}(g) \quad \Delta H=180 \mathrm{kJ} $$ Why are high temperatures needed to convert \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) to NO?

A biology experiment requires the preparation of a water bath at \(37.0^{\circ} \mathrm{C}\) (body temperature). The temperature of the cold tap water is \(22.0^{\circ} \mathrm{C},\) and the temperature of the hot tap water is \(55.0^{\circ} \mathrm{C} .\) If a student starts with 90.0 \(\mathrm{g}\) cold water, what mass of hot water must be added to reach \(37.0^{\circ} \mathrm{C} ?\)

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