Answer:
9.43 L^-2mol^-2
Explanation:
The ICE table was constructed and shown in the image attached. After construction of the ICE table, the equilibrium constant is easily determined from the equation of the reaction and law of active masses as shown in the image attached.
The equilibrium constant temperature is "9.43".
Initial concentration:
[tex]\to [CO] = 0.27 \ M\\\\[/tex]
[tex]\to [H_2] = 0.49\ M\\\\[/tex]
Using the ICE table for constructed as follows:
Equation:
[tex]CO(g) + 2H_2\ (g) \rightleftharpoons CH_3OH(g)\\\\[/tex]
Initial [tex]\ (M): \ \ \ \ \ \ \ \ \ 0.27 \ \ \ \ \ \ \ \ \ 0.49 \ \ \ \ \ \ \ \ \ 0\\\\[/tex]
Change[tex]\ (M): \ \ \ \ \ \ \ \ \ -x \ \ \ \ \ \ \ \ \ - 2x \ \ \ \ \ \ \ \ \ -x\\\\[/tex]
Equal [tex]\ (M) : \ \ \ \ \ \ \ \ \ 0.27-X \ \ \ \ \ \ \ \ \ 0.49-2x \ \ \ \ \ \ \ \ \ x \\\\[/tex]
[tex]\to [(CH_3OH)]_{equil} = (x) M\ = 0.11\ M\\\\ \to x= 0.11 \\\\[/tex]
Calculating the equilibrium concentration of [tex]H_2[/tex]:
[tex]\to [H_2] = (0.49-2x)\ M[/tex]
[tex]= (0.49-2\times 0.11)\ M \\\\ = (0.49- 0.22)\ M \\\\= 0.27\ M \\\\[/tex]
Calculating the equilibrium concentration of [tex]CO[/tex]:
[tex]\to [CO] = (0.27 - x)\ M[/tex]
[tex]= (0.27 -0.11) \ M \\\\ = 0.16\ M \\\\[/tex]
Calculating the equilibrium constant value:
[tex]\to K_c =\frac{[CH_3OH ]}{[CO][H_2]^2}[/tex]
[tex]=\frac{(0.11)}{(0.16)(0.27)^2} \\\\ =\frac{(0.11)}{(0.16)(0.0729)} \\\\ =\frac{(0.11)}{(0.011664)} \\\\= 9.43[/tex]
Therefore, the final answer is "9.43"
Find out more information about the temperature here:
brainly.com/question/15267055
