Question

1.Ammonia is created in the Haber process in a rigid container (nitrogen gas plus hydrogen gas react to form ammonia gas). 2.0 moles of hydrogen gas are mixed with 4.0 moles of nitrogen gas. The initial pressure exerted on the container is 5.0 atm. Assuming the reaction runs to completion, what will the pressure (in atm) on the vessel be after the reaction takes place?

2.Pentane gas reacts with oxygen gas to give carbon dioxide gas and water vapor (gas). If you mix pentane and oxygen in the correct stoichiometric ratio, and if the total pressure of the mixture is 180 mm Hg, what are the partial pressures of pentane ( mmHg) and oxygen ( mm Hg)? If the temperature and volume do not change, what is the pressure of the water vapor ( mm Hg) after reaction

can you please show your work,thank you!

Answer 1

1.

The reaction between nitrogen gas and hydrogen gas to produce ammonia gas is

N2g)3H2 2NH

It is being assumed that the reaction runs to completion after starting with 2.0 moles of Hydrogen gas and 4.0 moles of nitrogen gas.

According to the balanced reaction, 1 mole of N2 gas reacts with 3 moles of H2 gas to form 2 moles of ammonia gas.

Hence, given that

nH 2.0 mol

n N4.0 mol

The pressure exerted is 5.0 atm initially when the number of moles of gas in the container is

ninitial nN, +nH,= 4.0 mol+2.0 mol = 6.0 mol

assuming constant volume and temperature.

Initial pressure in the container is

Pinitial= 5.0 atm

Hence, assuming the gases behave ideally

initial RT Pinitial V V ninitial 6.0 mol 5.0 atm RT Pinitial

Note that V, T and R are all constants. Hence, the pressure in the container is only dependent on the number of moles of gas.

i.e pressure is directly proportional to the number of moles of gas. So we can write

n final Pfinal Minitial Pinitial

The number of moles of nitrogen gas that will react with 2.0 mol of H2 gas is

mol N2 x 2.0 mol H 0.67 mol N2 3 mol H

Hence, all 2.0 mol of H2 will react with 0.67 mol of N2 to form ammonia gas. Here H2 is the limiting reactant.

Note that 3 mol of H2 forms 2 mol NH3. Hence, the number of moles of ammonia gas formed from 2 mol H2 is

2 mol NH3 x 2.0 mol H2 1.33 mol NH3 3 mol H

Hence, the constituents in the chamber after completion is

nNH 1.33 mol

(completely consumed as it is the limiting reactant)

(nitrogen gas is in excess, hence, there is 3.33 mol leftover after completion)

Hence, the total number of moles of gas after completion of reaction is

final = 1.33 mol+0 mol+3.33 mol 4.66 mol

.

Hence, the final pressure in the container is

initial final Pinitial 6.0 mol Pfinal 4.66 mol 5.0 atm Pfinal 5.0 atm x 4.66 тol 3.9 atm Pfinal 6,0 mol

Hence, the pressure in the container at completion of reaction is about 3.9 atm.

2.

The balanced reaction between pentane and oxygen to produce water and CO2 is

$C_5H_{12}_{(g)} +8 O_2_{(g)} \rightarrow 5CO_2_{(g)} + 6H_2O_{(g)}$

It is given that pentane and oxygen are being mixed in correct stoichiometric ratio i.e. 1 mol of pentane per 8 mol of oxygen gas.

Hence, the mole fraction of pentane and oxygen in the mixture is

$\chi_{pentane} = \frac{1 \ mol}{1 \ mol + 8 \ mol} = \frac{1}{9}$

8 mol Xory gen 1 mol8 mol

Total pressure of the pentane and oxygen gas mixture,

Since the partial pressures of the gas is directly proportional to their mole fractions, the partial pressures are given by

$p_{O_2} = \chi_{oxygen} \times p_{total} = \frac{8}{9} \times 180 \ mm \ Hg = 160 \ mmHg$

$p_{C_5H_{12}} = \chi_{pentane} \times p_{total} = \frac{1}{9} \times 180 \ mm \ Hg = 20 \ mmHg$

Hence, the partial pressures of oxygen and pentane gas are 160 mmHg and 20 mmHg respectively.

After the reaction goes to completion at constant temperature and volume., we get 5 moles CO2 and 6 moles of H2O gases.(Note that since the reactant are taken in stoichiometric ratio, they are completely exhausted at completion)

We know that 9 moles of reactants resulted in a total pressure of 180 mmHg. Hence, 5 mol CO2 + 6 mol H2O =11 mol of products will result in a total pressure of

$p_{total} = \frac{180 \ mmHg}{9 \ mol} \times 11 \ mol = 220 \ mmHg$

mole fraction of H2O gas at completion is

$\chi_{H_2O} = \frac{6 \ mol \ H_2O}{5 \ mol \ CO_2 + \ 6 \ mol \ H_2O} = \frac{6}{11}$

Hence, the partial pressure of H2O vapor at completion can be calculated as

$p_{H_2O} = \chi_{H_2O} \times p_{total} = \frac{6}{11} \times 220 \ mmHg = 120 \ mmHg$

Hence, the partial pressure of H2O at completion of reaction is 120 mmHg.