the below 2questions i could not answer while revising for this chapter pls help me out a little.
58. When a vehicle is parked in the sunlight on a hot summer day, the temperature inside can approach 55°C. One company has patented a non-CFC propelled aerosol that can be sprayed inside a vehicle to reduce the temperature to 25°C within seconds. The spray contains a mixture of two liquids: 10% ethanol, C2H5OH, and 90% water by mass.
(a) Use thermochemical equations, and the corresponding enthalpy changes, to explain how the spray works.
(b) 1.0 g of the aerosol is sprayed into a hot vehicle. How much heat (in kJ) can be absorbed due to vaporization of the aerosol?
Note: ΔHvap of water = 44.0 kJ/mol and ΔHvap of ethanol = 38.56 kJ/mol
59. Consider methane, CH4, and hydrogen, H2, as possible fuel sources.
(a) Write the chemical equation for the complete combustion of each fuel. Then find the enthalpy of combustion, ΔHcomb, of each fuel. Express your answers in kJ/mol and kJ/g. Assume that water vapour, rather than liquid water, is formed in both reactions.
(b) Which is the more energetic fuel, per unit mass?
(c) Consider a fixed mass of each fuel. Which fuel would allow you to drive a greater distance? Explain briefly.
58. (a) To explain how the spray works, we need to consider the enthalpy changes associated with the vaporization of water and ethanol.
First, let's write the thermochemical equation for the vaporization of water:
H2O(l) -> H2O(g) ΔH = +44.0 kJ/mol
And the thermochemical equation for the vaporization of ethanol:
C2H5OH(l) -> C2H5OH(g) ΔH = +38.56 kJ/mol
When the spray is released inside the vehicle, the heat from the surrounding air is transferred to the liquid mixture. This heat is then used to vaporize the ethanol and water.
As the liquids vaporize, they absorb energy from their surroundings, including the air inside the vehicle. This energy absorbed causes the temperature inside the vehicle to decrease rapidly, providing the cooling effect.
(b) To calculate the heat absorbed due to the vaporization of the aerosol, we need to know the amount of heat absorbed per gram of the aerosol.
First, we need to calculate the moles of ethanol and water in 1 gram of the aerosol:
Mass of ethanol = 0.10 g
Molar mass of ethanol = 46.07 g/mol
Moles of ethanol = mass / molar mass = 0.10 g / 46.07 g/mol = 0.00217 mol
Mass of water = 0.90 g
Molar mass of water = 18.02 g/mol
Moles of water = mass / molar mass = 0.90 g / 18.02 g/mol = 0.0499 mol
Now, let's calculate the heat absorbed due to the vaporization of the aerosol:
Heat absorbed = (moles of ethanol * ΔHvap of ethanol) + (moles of water * ΔHvap of water)
Heat absorbed = (0.00217 mol * 38.56 kJ/mol) + (0.0499 mol * 44.0 kJ/mol)
Heat absorbed ≈ 0.084 kJ
Therefore, approximately 0.084 kJ of heat can be absorbed due to the vaporization of the aerosol.
59. (a) The complete combustion equations for methane and hydrogen are as follows:
Methane: CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g)
Enthalpy of combustion (ΔHcomb) of methane = -890.4 kJ/mol
Hydrogen: 2H2(g) + O2(g) -> 2H2O(g)
Enthalpy of combustion (ΔHcomb) of hydrogen = -483.6 kJ/mol
To find the enthalpy of combustion in kJ/g, we divide the ΔHcomb by the molar mass of the fuel:
Molar mass of methane (CH4) = 16.04 g/mol
Enthalpy of combustion of methane in kJ/g = -890.4 kJ/mol / 16.04 g/mol ≈ -55.6 kJ/g
Molar mass of hydrogen (H2) = 2.02 g/mol
Enthalpy of combustion of hydrogen in kJ/g = -483.6 kJ/mol / 2.02 g/mol ≈ -239.6 kJ/g
(b) The more energetic fuel per unit mass is hydrogen, as its enthalpy of combustion is higher (more negative) at approximately -239.6 kJ/g, compared to methane at approximately -55.6 kJ/g.
(c) For a fixed mass of each fuel, the fuel that allows you to drive a greater distance depends on the energy content per mass of the fuel and the fuel consumption rate of the engine being used.
Since hydrogen has a higher energy content per mass (-239.6 kJ/g) compared to methane (-55.6 kJ/g), it can potentially provide more energy and allow you to drive a greater distance for the same mass of fuel. However, hydrogen is also less dense than methane, and thus requires larger storage tanks or more frequent refueling.
The actual distance you can drive will also depend on the specific efficiency of the engine and the other factors such as vehicle weight, aerodynamics, and driving conditions.