Question

At equilibrium at 1200°C, [I2] = 9.5 ×10–2 M and [I] = 3.2 × 10–2 M. What is the value of Keq for this system?

Answer 1

B 1.1 × 10^–2

Step-by-step explanation:

just got it right on the test

Answer 2

Answer:The value of equilibrium constant is
`1.0778* 10^(-2)`.

Step-by-step explanation:

`I_2\rightleftharpoons 2I^-`

`[I_2]=9.5* 10^(-2) M,[I^-]=3.2* 10^(-2) M`

Expression of an equilibrium constant is given as:

`K_(eq)=([I^-]^2)/([I_2])=(3.2* 10^(-2)* 3.2* 10^(-2))/(9.5* 10^(-2))`

`K_(eq)=1.0778* 10^(-2)`

The value of equilibrium constant is
`1.0778* 10^(-2)`.