Question

Gaseous hydrogen and solid arsenic oxide react to form gaseous arsine and water by the following reaction:

6H₂ + As₂O₃ --> 2AsH₃ + 3H₂O

Intro a 4.0L reactor are placed excess As₂O₃ and 0.143 moles H₂

(A) the reactikon is carried out at 125 degrees C, a temperature at which all the water formed is gaseous. If all the hydrogen reacts, what is the pressure in the reactor after the reaction is completed?

(B) After the reaction is completed, the temperature is decreased to 25 degrees C, a temperature at which a portion of the water condenses. If all the hydrogen reacts, what is the mass of water that condenses? What is the pressure in the reactor after the reaction is completed and brought to this temperature?

(C) You conduct the reaction, but find that all the hydrogen does not react. After the reaction temperatrue is decreased to 25 degrees C, 0.90g of water has been found to condense. What is the fractional conversion of the reaction? What is the pressure in the reactor after the reaction is completed and brought to 25 degrees C.

Answer 1

(A)

Pressure of the reactor after completion of reaction.

Reaction is carried out at 125 C. All the hydrogen reacts, that means 0.143 moles. Volume is 4.0 L.

T = 125 C = 398.15 K

V = 4.0 L

All the hydrogen reacts with As2O3. 6 mole H2 is required to react with 1/6 mole of As2O3.

Moles of As2O3 = 0.143 * 1/6 = 0.02383 moles

n = total number of moles = 0.143 + 0.02383 = 0.1668 moles

Number of moles before and after reaction will stay the same.

Let’s calculate pressure

PV = nRT

P = nRT/V = 0.1668 * 8.314 * 10^-2 * 398.15 / 4 = 1.380 bar

(B)

Mass of water produced in reaction

6H2 + As2O3 à 2AsH3 + 3H2O

0.1668 moles of reactants produce same amount of products,0.1668 moles.

Around 3/5 moles are of water.

0.1668 * 3/5 = 0.10008 moles of water

Molar mass of water is 18.015.

Weight of water is 18.015 * 0.10008 = 1.8029 g of water

0.10008 moles were converted to water. And condensed water has negligible effect on pressure as compared to gases. That’s why we will subtract moles of water from calculation from pressure.

n = 0.1668 – 0.10008 = 0.06672 moles

Pressure in the reactor at decreased temperature. (25C)

P = nRT/V = 0.06672 * 8.314 * 10^-2 * 298.15 / 4 = 0.413 bar

(C)

Only 0.9 g of water was formed. That makes number of moles of water as 0.9/18.015=0.0499 moles

As we can see in reaction, half moles of water were formed as compared to H2.

Therefore, moles of H2 reacted = 0.0499 * 2 = 0.0998 moles of H2 reacted.

Fractional conversion = moles reacted/moles fed to reactor

                                       = 0.0998/0.143 = 0.698

We will need to subtract moles of water from total number of moles as condensed water has negligible effect on pressure.

n = 0.1668 – 0.0499 = 0.1169 moles

P = nRT/V = 0.1169 * 8.314 * 10^-2 * 298.15 / 4 = 0.724 bar