Methanol (CH,OH) can be synthesized by the following reaction: CO(g) + 2H2(g) CH
ID: 1026485 • Letter: M
Question
Methanol (CH,OH) can be synthesized by the following reaction: CO(g) + 2H2(g) CH3OH(g) A 2.00-L reaction vessel, initially at 152°C, contains carbon monoxide gas at a partial pressure of 454 Torr and hydrogen gas at a partial pressure of 518 Torr. a. What is the total pressure in the reaction vessel before the reaction? b. How many moles of each reactant are present before the reaction? c. What is the limiting reactant? d. How many moles of methanol are produced? What is the partial pres- sure of methanol? e. How many moles of the excess reactant are left after the reaction? What is the partial pressure of the excess reactant? f. What is the total pressure in the reaction vessel after the reaction?Explanation / Answer
a) The total pressure in the reaction vessel is the sum of the partial pressures of each of the reactants; hence, the total pressure is
Total pressure = (partial pressure of CO gas) + (partial pressure of H2 gas) = (454 + 518) torr = 972 torr (ans).
b) We have the pressure inside the container (before the reaction starts, i.e, no product is present) is P = 972 torr = (972 torr)*(1 atm/760 torr) = 1.2789 atm.
The volume of the vessel is V = 2.00 L.
The temperature of the reactants is 152°C, i.e, T = (152 + 273) K = 425 K.
Use the ideal gas law to determine the total number of mole(s) of gaseous reactants as
P*V = n*R*T where n = number of moles of gaseous reactants.
Plug in values and obtain
(1.2789 atm)*(2.00 L) = n*(0.082 L-atm/mol.K)*(425 K)
====> 2.5578 L-atm = n*(34.85 L-atm/mol)
====> n = (2.5578 L-atm)/(34.85 L-atm/mol) = 0.0734 mole.
Next use Dalton’s law of partial pressures to determine the mole fraction of each gas in the reactant mixture as
P(CO) = (mole fraction of CO)*(total pressure)
P(H2) = (mole fraction of H2)*(total pressure)
Plug in values and obtain
454 torr = (mole fraction of CO)*(972 torr)
=====> mole fraction of CO = (454 torr)/(972 torr) = 0.4670 …..(1)
518 torr = (mole fraction of H2)*(972 torr)
=====> mole fraction of H2 = (518 torr)/(972 torr) = 0.5330 …..(2)
Again, (mole fraction of CO) + (mole fraction of H2) = 1
======> (moles of CO)/(total number of moles of gaseous reactants) + (moles of H2/total number of moles of gaseous reactants) = 1
======> n(CO)/(0.0734 mole) + n(H2)/(0.0734 mole) = 1
======> n(CO) + n(H2) = 0.0734 ……(3)
Again, n(CO)/(0.0734 mole) = 0.4670
=====> n(CO) = (0.4670)*(0.0734 mole) = 0.0343 mole (ans).
n(H2)/(0.0734 mole) = 0.5330
======> n(H2) = (0.5330)*(0.0734 mole) = 0.0391 mole (ans).
Note that n(CO) + n(H2) = (0.0343 + 0.0391) mole = 0.0734 mole.
c) As per the stoichiometric equation,
1 mole CO = 2 moles H2 = 1 mole CH3OH
Therefore, 0.0343 mole CO = (0.0343 mole CO)*(2 moles H2/1 mole CO) = 0.0686 mole H2.
0.0391 mole H2 = (0.0391 mole H2)*(1 mole CO/2 mole CO) = 0.01955 mole CO.
Obviously, we do not have 0.0686 mole H2 (to react with 0.0343 mole CO), but we have more than 0.01955 mole CO (to react with 0.0391 mole H2). Therefore, H2 is the limiting reactant (ans).
d) The yield of the product is decided by the limiting reactant.
Mole(s) of CH3OH = (0.0391 mole H2)*(1 mole CH3OH/2 moles H2) = 0.01955 mole CH3OH (ans).
The total pressure of the system remains constant and hence, the total number of gaseous products and reactants must remain constant; therefore, partial pressure of CH3OH = (moles fraction of CH3OH)*(total pressure) = (0.01955 mole/0.0734 mole)*(972 torr) = 258.89 torr 259 torr (ans).