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A mixture of gaseous CO and H2, called synthesis gas, is used commercially to prepare methanol (CH3OH), a compound considered an alternative fuel to gasoline. Under equilibrium conditions at 540.2 K, [H2] = 0.067 mol [CO]=0.021 mol/L and [CH3OH]=0.040 mol/L what is the value of Kc for this reaction at 505.0 K?

Respuesta :

Answer : The value of equilibrium constant (K) is, 424.3

Explanation :  Given,

Concentration of [tex]H_2[/tex] at equilibrium = 0.067 mol

Concentration of [tex]CO[/tex] at equilibrium = 0.021 mol

Concentration of [tex]CH_3OH[/tex] at equilibrium = 0.040 mol

The given chemical reaction is:

[tex]CO+2H_2\rightarrow CH_3OH[/tex]

The expression for equilibrium constant is:

[tex]K_c=\frac{[CH_3OH]}{[CO][H_2]^2}[/tex]

Now put all the given values in this expression, we get:

[tex]K_c=\frac{(0.040)}{(0.021)\times (0.067)^2}[/tex]

[tex]K_c=424.3[/tex]

Thus, the value of equilibrium constant (K) is, 424.3

Histrionicus

The value of Kc for this reaction at 505.0 K would be - 373.3.

Given:

[tex][H_2] = 0.063\ M\\\\[/tex]

[CO]=0.021 mol/L

[tex][CH_3OH] = 0.040\ M\\[/tex]

Reaction:

[tex]CO + 2H_2 \rightarrow CH_3OH[/tex]

Solution:

  • As we know the formula,

[tex]Kc = \frac{[CH_3OH]}{[CO][H_2]^2}[/tex]

  • By substituting the values, we get

Kc= [tex]\frac{{[0.040]}}{0.027\times (0.063)^2}[/tex]

=  373.3

The value of Kc for this reaction at 505.0 K would be - 373.3.

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