Respuesta :
The question is incomplete, here is the complete question:
Consider the reaction below.
0.0067 M SbCl₅ is placed in a one liter flask and the system is allowed to reach equilibrium. What is final concentration of SbCl₅? Kc at this temperature is 2.5 x 10-2.
[tex]SbCl_5(g)\rightleftharpoons SbCl_3(g)+Cl_2(g)[/tex]
Answer: The final concentration of [tex]SbCl_5[/tex] is 0.0012 M
Explanation:
We are given:
Initial concentration of [tex]SbCl_5[/tex] = 0.0067 M
For the given chemical equation:
[tex]SbCl_5(g)\rightleftharpoons SbCl_3(g)+Cl_2(g)[/tex]
Initial: 0.0067
At eqllm: 0.0067-x x x
The expression of [tex]K_c[/tex] for above equation follows:
[tex]K_c=\frac{[Cl_2][SbCl_3]}{[SbCl_5]}[/tex]
We are given:
[tex]K_c=2.5\times 10^{-2}[/tex]
Putting values in above expression, we get:
[tex]2.5\times 10^{-2}=\frac{x\times x}{(0.0067-x)}\\\\x=-0.0305,0.0055[/tex]
Neglecting the negative value of 'x' because concentration cannot be negative
So, final concentration of [tex]SbCl_5=(0.0067-x)=(0.0067-0.0055)=0.0012M[/tex]
Hence, the final concentration of [tex]SbCl_5[/tex] is 0.0012 M
