Respuesta :
Answer: The equilibrium concentration of [tex]CH_4\text{ and }CCl_4[/tex] is 0.377 M and equilibrium concentration of [tex]CH_2Cl_2[/tex] is 0.116 M
Explanation:
To calculate the molarity of solution, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]
- For Methane:
Volume of solution = 1 L
Moles of methane = 0.377 moles
Putting values in above equation, we get:
[tex]\text{Molarity of }CH_4=\frac{0.377mol}{1L}\\\\\text{Molarity of }CH_4=0.377M[/tex]
- For carbon tetrachloride:
Volume of solution = 1 L
Moles of carbon tetrachloride = 0.377 moles
Putting values in above equation, we get:
[tex]\text{Molarity of carbon tetrachloride}=\frac{0.377mol}{1L}\\\\\text{Molarity of carbon tetrachloride}=0.377M[/tex]
For the given chemical equation:
[tex]CH_4(g)+CCl_4(g)\rightarrow 2CH_2Cl_2(g)[/tex]
The expression for [tex]K_c[/tex] for the given equation follows:
[tex]K_c=\frac{[CH_2Cl_2]^2}{[CH_4][CCl_4]}[/tex]
We are given:
[tex]K_c=9.52\times 10^{-2}\\[CH_4]=0.377M\\[CCl_4]=0.377M[/tex]
Putting values in above equation, we get:
[tex]9.52\times 10^{-2}=\frac{[CH_2Cl_2]^2}{(0.377)\times (0.377)}[/tex]
[tex][CH_2Cl_2]=0.116M[/tex]
Hence, the equilibrium concentration of [tex]CH_4\text{ and }CCl_4[/tex] is 0.377 M and equilibrium concentration of [tex]CH_2Cl_2[/tex] is 0.116 M
