Chemistry, 06.02.2021 03:40 addisonrausch

A reaction mixture in a 5.19 L flask at a certain temperature contains 26.9 g CO and 2.34 g H2. At equilibrium, the flask contains 8.65 g CH3OH. Calculate the equilibrium constant (Kc) for the reaction at this temperature.

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Answer from: austinlogan3218

26.7

Explanation:

The reaction that takes place is:

CO + 2H₂ ↔ CH₃OH

We convert the given masses to moles, using their respective molar masses:

CO ⇒ 26.9 g ÷ 28 g/mol = 0.961 molH₂ ⇒ 2.34 g ÷ 2 g/mol = 1.17 molCH₃OH ⇒ 8.65 g ÷ 32 g/mol = 0.270 mol

The initial concentrations for each species are:

CO ⇒ 0.961 mol / 5.19 L = 0.185 MH₂ ⇒ 1.17 mol / 5.19 L = 0.225 MCH₃OH ⇒ 0

While the equilibrium concentration for CH₃OH, [CH₃OH]eq is:

0.270 mol / 5.19 L = 0.052 M

We put the data in a table:

CO + 2H₂ ↔ CH₃OH

initial 0.185 0.225 ↔ 0

eq (0.185 - x) (0.225-2x) ↔ x

We know that x = 0.052 M (That's the equilibrium concentration of CH₃OH).

We proceed to calculate [CO]eq and [H₂]eq:

[CO]eq = 0.185 - 0.052 = 0.133 M[H₂]eq = 0.225 - 2*0.052 = 0.121 M

Finally we calculate the equilibrium constant:

Kc = $\frac{[CH_3OH]_{eq}}{[CO]_{eq}([H_2]_{eq})^2}$ = 26.7

Answer from: stacy021603

The equilibrium constant (Kc) for the reaction at this temperature is 26.7

Chemical reaction:

CO + 2H₂ ⇄ CH₃OH

Conversion of moles:

CO ⇒ $\frac{26.9 g}{28 g/mol} = 0.961 mol$

H₂ ⇒ $\frac{2.34 g }{ 2 g/mol} =1.17 mol$

CH₃OH ⇒ $\frac{8.65 g}{32 g/mol} = 0.270 mol$

The initial concentrations for each species are:

CO ⇒ $\frac{0.961 mol}{5.19 L} = 0.185 M$

H₂ ⇒ $\frac{1.17 mol}{5.19 L } = 0.225 M$

CH₃OH ⇒ 0

While the equilibrium concentration for CH₃OH, [CH₃OH]eq is:

$\frac{0.270 mol}{ 5.19 L} = 0.052 M$

We put the data in a table:

CO + 2H₂ ⇄ CH₃OH

Initial 0.185 0.225 0

Equilibrium (0.185 - x) (0.225-2x) x

x = 0.052 M (That's the equilibrium concentration of CH₃OH).

We proceed to calculate [CO]eq and [H₂]eq:

$[CO]_{eq}= 0.185 - 0.052 = 0.133 M\\\\\\[H_2]_{eq} = 0.225 - 2*0.052 = 0.121 M$

Calculation of equilibrium constant:

$K_c =\frac{[CH_3OH]_{eq}}{[CO]_{eq}[H_2_{eq}]^2} = 26.7$

Thus, the equilibrium constant (Kc) for the reaction at this temperature is 26.7

Find more information about Equilibrium constant here:

Answer from: Quest

wher is the rest of the question

Answer from: Quest

Ithink ahopefully im correct

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A reaction mixture in a 5.19 L flask at a certain temperature contains 26.9 g CO and 2.34 g H2. At e...

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