Question

Consider the reaction below. Initially the concentration of SO2Cl2 is 0.1000 M. Solve for the equilibrium concentration of SO2Cl2((g). SO2Cl2(g) ←⎯⎯→ SO2(g) + Cl2(g) Kc = 2.99 x 10-7 at 227 °C

Answer 1

`[SO_2Cl_2] = 0.09983 M`

Step-by-step explanation:

Write the balance chemical equation ,

`SO_2Cl_2((g) = SO_2(g) + Cl_2(g)`

initial concenration of
`SO_2Cl_2((g) =0.1M`

lets assume that degree of dissociation=
`\alpha`

concenration of each component at equilibrium:

`[SO_2Cl_2] = 0.1-0.1\alpha`

`[SO_2] = 0.1\alpha`

`[Cl_2] = 0.1\alpha`

`Kc =(0.1\alpha * 0.1\alpha)/(0.1-0.1\alpha)`

`Kc =(0.1\alpha * \alpha)/(1-\alpha)`

as
`\alpha` is very small then we can neglect
`1-\alpha`

therefore ,

`Kc ={0.1\alpha * \alpha}`

`\alpha =\sqrt{(Kc)/(0.1)}`

`\alpha = 1.73 * 10^(-3)`

Eqilibrium concenration of
`[SO_2Cl_2] = 0.1-0.1\alpha = 0.1-0.1* 0.00173`

`[SO_2Cl_2] = 0.09983 M`