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

Answer:The value of equilibrium constant is $1.0778\times 10^{-2}$.

Explanation:

$I_2\rightleftharpoons 2I^-$

$[I_2]=9.5\times 10^{-2} M,[I^-]=3.2\times 10^{-2} M$

Expression of an equilibrium constant is given as:

$K_{eq}=\frac{[I^-]^2}{[I_2]}=\frac{3.2\times 10^{-2}\times 3.2\times 10^{-2}}{9.5\times 10^{-2}}$

$K_{eq}=1.0778\times 10^{-2}$

The value of equilibrium constant is $1.0778\times 10^{-2}$.