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aleksandrvk [35]
3 years ago
10

Consider the reaction below. Initially the concentration of SO2Cl2 is 0.1000 M. Solve for the equilibrium concentration of SO2Cl

2((g). SO2Cl2(g) ←⎯⎯→ SO2(g) + Cl2(g) Kc = 2.99 x 10-7 at 227 °C

Chemistry
1 answer:
tensa zangetsu [6.8K]3 years ago
4 0

Answer:

[SO_2Cl_2] = 0.09983 M

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 =\frac{0.1\alpha \times 0.1\alpha}{0.1-0.1\alpha}

Kc =\frac{0.1\alpha \times \alpha}{1-\alpha}

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

therefore ,

Kc ={0.1\alpha \times \alpha}

\alpha =\sqrt{\frac{Kc}{0.1}}

\alpha = 1.73 \times 10^{-3}

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

[SO_2Cl_2] = 0.09983 M

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