Given that the equilibrium constant for the reaction, 2 SO_2(g)+ O_2(g) ⇌ 2 SO_3(g) has a value of 278 at a particular temperature. What is the value…

$$2 \mathrm{SO}_2(g)+\mathrm{O}_2(g) \rightleftharpoons 2 \mathrm{SO}_3(g)$$2SO2​(g)+O2​(g)⇌2SO3​(g) Equilibrium constant for this reaction, $$\begin{gathered} K=\frac{\left[\mathrm{SO}_3\right]^2}{\left[\mathrm{SO}_2\right]^2\left[\mathrm{O}_2\right]} \\ \mathrm{SO}_3(g) \rightleftharpoons \mathrm{SO}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \end{gathered}$$K=[SO2​]2[O2​][SO3​]2​SO3​(g)⇌SO2​(g)+21​O2​(g)​ Equilibrium constant for this reaction, $$K^{\prime}=\frac{\left[\mathrm{SO}_2\right]\left[\mathrm{O}_2\right]^{\frac{1}{2}}}{\left[\mathrm{SO}_3\right]}$$K′=[SO3​][SO2​][O2​]21​​ On squaring Eq. (ii) both sides, we have $$\left(K^{\prime}\right)^2=\frac{\left[\mathrm{SO}_2\right]^2\left[\mathrm{O}_2\right]}{\left[\mathrm{SO}_3\right]^2}$$(K′)2=[SO3​]2[SO2​]2[O2​]​ $$\begin{aligned} & =\frac{1}{K} \\ & =\frac{1}{278} \\ \therefore \quad K^{\prime} & =\sqrt{\frac{1}{278}} \\ & =\sqrt{0.003597} \\ & =5.99 \times 10^{-2} \\ & \approx 6 \times 10^{-2}\end{aligned}$$∴K′​=K1​=2781​=2781​​=0.003597​=5.99×10−2≈6×10−2​