Question

a) A 1.500 L flask is filled with a mixture of 1.20 g H2 and 8.40 g O2, at 25°C. The mixture is ignited, and hydrogen and oxygen combine to form water.

What is the total pressure inside the flask before the reaction? (Report your answer in units of atmospheres.)

What is the total pressure after the reaction, once the flask is returned to 25°C? (Vapour pressure of water at 25°C is 23.8 mm Hg.)

b) In the photoelectric effect, the work function is the energy that must be supplied to overcome the attractive forces that hold an electron in the metal. For mercury, the work function is equal to 435 kJ mol-1 of photons. What is the kinetic energy, in joules, of the ejected electrons when light with a wavelength of 220 nm strikes the surface of mercury?

Answer 1

Given :

Volume of flask = 1.500 L

Mass of H2 = 1.20 g , 8.40 g O2 At 25 deg C = 25deg C + 273.15 = 298.15 K

We calculate moles of each gas

Number of moles of H2 = 1.20 g / 2.0158 g per mol = 0.5953 mol

Mol of O2 = 8.40 g /31.998 g per mol

= 0.2625 mol O2

Total moles = mol H2 + mol O2

= 0.5953 mol + 0.2625 mol

=0.8578 mol of mixture

Total pressure = nRT / V ( ideal gas law)

n : number of moles , R is gas constant = 0.08206 L atm per K per mol

T is in K , and V is in L

Total pressure = 0.8578 mol * 0.08206 L atm per K per mol * 298.15 K / 1.500 L

= 13.99 atm

14.0 atm

Total pressure of the gas inside = 14.0 atm

To find total pressure after reaction we use the reaction

H2 (g) + ½ O2 (g) -- > H2O (l)

Lets find limiting reactant in above reaction:

Number of moles of H2O from mole of H2

Moles of H2 = moles of H2 * 1 mol H2O / 1 mol H2

= 0.5953 mol H2O

Moles of H2O from moles of O2

= moles of O2 * 1 mol H2O / ½ moles of O2

= 0.2625 mol * 1 mol H2O / ½ mol O2

= 0.5250 mol H2O

Moles of O2 are limiting reactant

Moles of O2 will be completely consumed

We calculate number of moles of H2 remained

Number of moles of H2 remained = 0.5953 mol H2 – 0.2652 mol O2 * 1 mol H1 / ½ mol O2

= 0.070 mol H2

Now when temperature is returned to 25 deg C then all the H2O is in liquid form so,

Pressure of H2 on liquid = (0.070 mol * 0.08206 L atm per K per mol * 298.15 K ) / 1.500 L

= 1.15 atm

At 25 deg C pressure of H2O is 23.8 mmHg = 23.8 mmHg * 1 atm / 760 mmHg =0.03132 atm

Total pressure inside the tank = 1.15 atm + 0.03132 atm = 1.18 atm

Question 2)

Work function for mercury = 435 kJ per mol

Wavelength of light = 220 nm

This problem is related to photoelectric effect

Energy of striking photon = Work function + kinetic energy of ejected electron

Lets convert given wavelength into m

l = 220 nm * 1E-9 m / 1 nm = 2.20 E -7 m

We know Energy of =h c / l

Here c is speed of light ( 3.0E8 m per s ) , h is planks constant = 6.626E-34 Js )

Lets calculate energy of striking photon

Energy = 6.626 E-34 Js * 3.0 E8 m per s / 2.20 E-7 m

= 9.03 E-19 J

Now we convert work function into J per photon

Work function in J /photon = ( 435 kJ/ mol ) * (1000 J / kJ ) * (1 mol /6.02 E23 photon)

= 7.22 E -19 J

Energy of striking photon = Work function + kinetic energy of ejected electron

Kinetic energy = 9.03E-19 J – 7.22 E-19 J = 1.81 E -19 J

Answer: Kinetic energy of ejected electron = 1.81 E-19 J