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Sagot :
Answer:
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 mol
- H₂ ⇒ 2.34 g ÷ 2 g/mol = 1.17 mol
- CH₃OH ⇒ 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 M
- H₂ ⇒ 1.17 mol / 5.19 L = 0.225 M
- CH₃OH ⇒ 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 = [tex]\frac{[CH_3OH]_{eq}}{[CO]_{eq}([H_2]_{eq})^2}[/tex] = 26.7
The equilibrium constant (Kc) for the reaction at this temperature is 26.7
Chemical reaction:
CO + 2H₂ ⇄ CH₃OH
- Conversion of moles:
CO ⇒ [tex]\frac{26.9 g}{28 g/mol} = 0.961 mol[/tex]
H₂ ⇒ [tex]\frac{2.34 g }{ 2 g/mol} =1.17 mol[/tex]
CH₃OH ⇒ [tex]\frac{8.65 g}{32 g/mol} = 0.270 mol[/tex]
- The initial concentrations for each species are:
CO ⇒ [tex]\frac{0.961 mol}{5.19 L} = 0.185 M[/tex]
H₂ ⇒ [tex]\frac{1.17 mol}{5.19 L } = 0.225 M[/tex]
CH₃OH ⇒ 0
- While the equilibrium concentration for CH₃OH, [CH₃OH]eq is:
[tex]\frac{0.270 mol}{ 5.19 L} = 0.052 M[/tex]
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:
[tex][CO]_{eq}= 0.185 - 0.052 = 0.133 M\\\\\\[H_2]_{eq} = 0.225 - 2*0.052 = 0.121 M[/tex]
- Calculation of equilibrium constant:
[tex]K_c =\frac{[CH_3OH]_{eq}}{[CO]_{eq}[H_2_{eq}]^2} = 26.7[/tex]
Thus, the equilibrium constant (Kc) for the reaction at this temperature is 26.7
Find more information about Equilibrium constant here:
brainly.com/question/12270624
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