Question

CO(g)+2H2(g)⇌CH3OH(g) A reaction mixture in a 5.23-L flask at a certain temperature initially contains 27.2 g CO and 2.36 g H2. At equilibrium, the flask contains 8.64 g CH3OH

Answer 1

Answer: 5.70M

Explanation:

Molar mass of CO = 28.01 g/mol

Molar mass of H2 = 2.02 g/mol

Molar mass of CH3OH = 32.05 g/mol.

To determine the amount of each compound in the reaction mixture we use the formula.

Amount in mol = reacting mass/molar mass.

Inputing the given values we have,

26.6 g CO x (1 mol / 28.01 g ) = 0.9496608354 mol of CO.

To calculate the concentration of CO we use C=n/v, where n=amount and v= volume of CO.

Inputing the values in the formula

[CO] = 0.9496608354 mol CO / 5.23 L = 0.18158 M CO

Repeating thesame procedure for H

Amount of H=2.36 g H2 x ( 1 mol / 2.02 g ) = 1.168316832 mol of H2

Concentration of H2 in the mixture

[H2] = 1.168316832 mol H2 / 5.23 L = 0.223388 M of H2

Amount of CH3OH is determine similarly using rmass/molar mass

8.66 CH3OH x (1 mol / 32.05 g ) = 0.2702028081 mol of CH3OH

Concentration of

[CH3OH] = 0.2702028081 mol CH3OH/ 5.23 L = 0.051664 M CH3OH

Now equilibrium constant is determined by

Kc = [CH3OH] / [CO] [H2]^2

=0.051664/0.18158×0.223388×0.223388.

=5.70