4. given the following equilibrium constants:\nn₂(g) + ½ o₂(g) ⇌ n₂o(g) kc = 2.7×10⁻¹⁸\nn₂o₄(g) ⇌ 2 no₂(g)…

4. given the following equilibrium constants:\nn₂(g) + ½ o₂(g) ⇌ n₂o(g) kc = 2.7×10⁻¹⁸\nn₂o₄(g) ⇌ 2 no₂(g) kc = 4.6×10⁻³\n½ n₂(g) + o₂(g) ⇌ no₂(g) kc = 4.1×10⁻⁹\ndetermine kc for\n2 n₂o(g) + 3 o₂(g) ⇌ 2 n₂o₄(g)\n(assume all constants are for the same temperature.)

4. given the following equilibrium constants:\nn₂(g) + ½ o₂(g) ⇌ n₂o(g) kc = 2.7×10⁻¹⁸\nn₂o₄(g) ⇌ 2 no₂(g) kc = 4.6×10⁻³\n½ n₂(g) + o₂(g) ⇌ no₂(g) kc = 4.1×10⁻⁹\ndetermine kc for\n2 n₂o(g) + 3 o₂(g) ⇌ 2 n₂o₄(g)\n(assume all constants are for the same temperature.)

Answer

Explanation:

Step1: Reverse the first reaction

When we reverse the reaction $N_2(g)+\frac{1}{2}O_2(g)\rightleftharpoons N_2O(g)$ with $K_{c1} = 2.7\times10^{-18}$, the new equilibrium - constant $K_{c1}'=\frac{1}{K_{c1}}=\frac{1}{2.7\times10^{-18}}$. The reversed reaction is $N_2O(g)\rightleftharpoons N_2(g)+\frac{1}{2}O_2(g)$.

Step2: Multiply the first - reversed reaction by 2

If we multiply the reaction $N_2O(g)\rightleftharpoons N_2(g)+\frac{1}{2}O_2(g)$ by 2, the new equilibrium - constant $K_{c1}''=(K_{c1}')^2=\left(\frac{1}{2.7\times10^{-18}}\right)^2$. The new reaction is $2N_2O(g)\rightleftharpoons 2N_2(g)+O_2(g)$.

Step3: Reverse the second reaction

When we reverse the reaction $N_2O_4(g)\rightleftharpoons 2NO_2(g)$ with $K_{c2} = 4.6\times10^{-3}$, the new equilibrium - constant $K_{c2}'=\frac{1}{K_{c2}}=\frac{1}{4.6\times10^{-3}}$. The reversed reaction is $2NO_2(g)\rightleftharpoons N_2O_4(g)$.

Step4: Multiply the third reaction by 2

If we multiply the reaction $\frac{1}{2}N_2(g)+O_2(g)\rightleftharpoons NO_2(g)$ with $K_{c3} = 4.1\times10^{-9}$ by 2, the new equilibrium - constant $K_{c3}''=(K_{c3})^2=(4.1\times10^{-9})^2$. The new reaction is $N_2(g)+2O_2(g)\rightleftharpoons 2NO_2(g)$.

Step5: Add the three modified reactions

Adding $2N_2O(g)\rightleftharpoons 2N_2(g)+O_2(g)$ ($K_{c1}''$), $2NO_2(g)\rightleftharpoons N_2O_4(g)$ ($K_{c2}'$), and $N_2(g)+2O_2(g)\rightleftharpoons 2NO_2(g)$ ($K_{c3}''$) gives $2N_2O(g)+3O_2(g)\rightleftharpoons 2N_2O_4(g)$. The overall equilibrium constant $K_{c}$ for the reaction $2N_2O(g)+3O_2(g)\rightleftharpoons 2N_2O_4(g)$ is the product of the equilibrium constants of the three modified reactions. [ \begin{align*} K_{c}&=K_{c1}''\times K_{c2}'\times K_{c3}''\ &=\left(\frac{1}{2.7\times10^{-18}}\right)^2\times\frac{1}{4.6\times10^{-3}}\times(4.1\times10^{-9})^2\ &=\frac{1^2\times1\times(4.1\times10^{-9})^2}{(2.7\times10^{-18})^2\times4.6\times10^{-3}}\ &=\frac{16.81\times10^{-18}}{(7.29\times10^{-36})\times4.6\times10^{-3}}\ &=\frac{16.81\times10^{-18}}{33.534\times10^{-39}}\ &=\frac{16.81}{33.534}\times10^{21}\ &\approx0.5\times10^{21}\ & = 5\times10^{20} \end{align*} ]

Answer:

$5\times 10^{20}$