The so-called hydrogen economy is based on hydrogen produced from water using solar energy. The gas is then burned as a fuel: 2H2(g) + O2(g) → 2H2O(l) A primary advantage of hydrogen as a fuel is that it is nonpolluting. A major disadvantage is that it is a gas and therefore is harder to store than liquids or solids. Calculate the volume of hydrogen gas at 28°C and 1.00 atm required to produce an amount of energy equivalent to that produced by the combustion of a gallon of octane (C8H18). The density of octane is 2.66 kg/gal, and its standard enthalpy of formation is −249.9 kJ/mol. Assume that the products of the combustion of octane are CO2(g) and H2O(l).

The energy of the reaction (Hr) is the sum of the enthalpies of the products (multiplied by their coefficients) less the sum of the enthalpies of the reactants (multiplied by their coefficients).

Hr = [(8*(-393) + 9*(-286)] - (-249.9)

Hr = - 5468.1 kJ/mol

For 1 gallon, the mass od octane must be:

m = 1*2.66 = 2.66 kg = 2660 g

The molar mass of octane is : 8*12 g/mol of C + 18*1 g/mol of H = 114 g/mol, and the number of moles:

n = 2660/114 = 23.33 moles

So, the energy produced is: - 5468.1 *23.33 = -127.6 x10³ kJ

The reaction of combustion of hydrogen:

2H₂(g) + O₂(g) → H₂O (l), entaphy of formation of H₂ = 0

Hr = -286 kJ/mol

To produce the same quantity of energy, will be necessary:

-286*n = -127.6 x10³

n = 446 moles of H₂

Because Hr is calculated by 2 moles of H₂, n = 446/2 = 223 moles

For the ideal gas law:

PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (0.082 atm*L/mol*K), and T is the temperature (28ºC + 273 = 301 K).

1*V = 223*0.082*301

V = 5,504 L

1 L = 0.264172 gallons

V = 1.5 x 10³ gallons of H₂

thomasdecabooter thomasdecabooter · Answer 2 · 2020-04-19T02:35:23+02:00

Answer:

The volume of hydrogen is 5075.9 L

Explanation:

Step 1: Data given

Temperature = 28.0 °C

Pressure = 1.00 atm

Enthalpy of formation for octane is -249.95 kJ/mol

Enthalpy of formation for CO2 is -393.5 kJ/mol

Enthalpy of formation for liquid water is -285.8 kJ/mol

The standard enthalpy of formation is −249.9 kJ/mol.

Step 2: The balanced equation

C8H18 + 12/2O2 →9H2O + 8 CO2

Step 3: Calculate ΔH° for the reaction

ΔH° = 9*(-241.8/ mol) + 8 *(-393.5 kj/mol) - 1*(-249.9 kj/mol)

ΔH° = -5074 kJ/mol

Step 4: Calculate moles octane

Moles octane = mass octane / molar mass octane

Moles octane = 2660 grams / 114.23 g/mol

Moles octane = 23.2 moles

Step 5: Calculate heat produced by combustion of 23.2 moles of octane

Heat = 23.2 moles * 5074kJ/mol

Heat = 117716.8 kJ

Step 6: Calculate moles H2

The reaction of combustion of hydrogen:

2H2(g) + O2(g) → H2O (l)

ΔH = -286 kJ/mol

To produce the same amount of energy we need:

-286*n = -117716.8

n = 411 moles of H2

Since we calculate this for 2 moles of H2 we have to divide this by 2.

n = 411/2 = 205.5 moles H2

Step 7: Calculate volume of hydrogen

p*V = n *R*T

⇒with p = the pressure of the gas = 1.00 atm

⇒with V = the volume of hydrogen gas = TO BE DETERMINED

⇒with n = the moles of hydrogen = 205.5 moles

⇒with R = the gas constant = 0.08206 L*atm/mol * K

⇒with T = the temperature = 28 °C = 301 K

V = (n*R*T) / p

V = (205.5 * 0.08206 * 301) .1.00

V = 12023 liters = 1.2 * 10^4 L

The volume of hydrogen is 5075.9 L