Solveeit Logo
Login

Question

Chemistry Question on Law Of Chemical Equilibrium And Equilibrium Constant

The value of KcK_c for the reaction 3O2 (g)↔2O3 (g)3O_2 \,(g) \leftrightarrow 2O_3 \,(g) is 2.0×10–502.0 ×10^{–50} at 25°C. If the equilibrium concentration of O2O_2 in air at 25°C is 1.6×10–21.6 ×10^{–2}, what is the concentration of O3O_3?

Answer

The given reaction is : 3O2 (g)↔2O3 (g)3O_2 \,(g) \leftrightarrow 2O_3 \,(g)
Then, KcK_c = [O3(g)]2[O2(g)]3\frac{\bigg[O_3(g)\bigg]^2 }{ \bigg[O_2(g)\bigg]^3}
It is given that KcK_c = 2.0×10−502.0 × 10 ^{- 50} and [O2(g)]=1.6×10−2\big[O_2(g)\big]=1.6\times 10^{-2}
Then, we have,
2.0×10−502.0\times 10^{-50} = [O3(g)]2[1.6×10−2]3\frac{\bigg[O_3(g)\bigg]^2 }{ \bigg[1.6 × 10^{ - 2}\bigg]^3}

⇒\Rightarrow [O3(g)]2 [O_3(g)]^2 = 2.0×10−50×(1.6×10−2)32.0 × 10 ^{- 50} × (1.6 × 10 ^{- 2})^3
⇒\Rightarrow [O3(g)]2 [O_3(g)]^2 = 8.192×10−568.192 × 10 ^{-56}
⇒\Rightarrow [O3(g)] [O_3(g)] = 2.86×10−282.86 × 10 ^{- 28} MM

Hence, the concentration of O3O_3 is 2.86×10−282.86 × 10 ^{- 28} MM.