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: Manipulate the given reactions

We have the following reactions and their $K_c$ values:

  1. $N_2(g)+\frac{1}{2}O_2(g)\rightleftharpoons N_2O(g)$ with $K_{c1}=2.7\times 10^{-18}$
  2. $N_2O_4(g)\rightleftharpoons 2NO_2(g)$ with $K_{c2}=4.6\times 10^{-3}$
  3. $\frac{1}{2}N_2(g)+O_2(g)\rightleftharpoons NO_2(g)$ with $K_{c3}=4.1\times 10^{9}$

We want to get $2N_2O(g)+3O_2(g)\rightleftharpoons 2N_2O_4(g)$

First, reverse reaction (1) and multiply it by 2: $2N_2O(g)\rightleftharpoons 2N_2(g)+O_2(g)$ with $K_{c1}'=\left(\frac{1}{K_{c1}}\right)^2$

Second, reverse reaction (2) and multiply it by 2: $4NO_2(g)\rightleftharpoons 2N_2O_4(g)$ with $K_{c2}'=\left(\frac{1}{K_{c2}}\right)^2$

Third, multiply reaction (3) by 4: $2N_2(g)+4O_2(g)\rightleftharpoons 4NO_2(g)$ with $K_{c3}'=(K_{c3})^4$

Step2: Calculate the overall $K_c$

When we add these three - manipulated reactions together, the intermediate species ($N_2$ and $NO_2$) will cancel out.

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 manipulated reactions.

$K_c = K_{c1}'\times K_{c2}'\times K_{c3}'$

$K_{c1}'=\left(\frac{1}{2.7\times 10^{-18}}\right)^2=\frac{1}{(2.7\times 10^{-18})^2}$

$K_{c2}'=\left(\frac{1}{4.6\times 10^{-3}}\right)^2=\frac{1}{(4.6\times 10^{-3})^2}$

$K_{c3}'=(4.1\times 10^{9})^4$

$K_c=\frac{(4.1\times 10^{9})^4}{(2.7\times 10^{-18})^2\times(4.6\times 10^{-3})^2}$

$(4.1\times 10^{9})^4 = 4.1^4\times10^{36}=282.5761\times 10^{36}$

$(2.7\times 10^{-18})^2=2.7^2\times 10^{- 36}=7.29\times 10^{-36}$

$(4.6\times 10^{-3})^2=4.6^2\times 10^{-6}=21.16\times 10^{-6}$

$K_c=\frac{282.5761\times 10^{36}}{7.29\times 10^{-36}\times21.16\times 10^{-6}}$

$K_c=\frac{282.5761\times 10^{36}}{(7.29\times21.16)\times 10^{-42}}$

$7.29\times21.16 = 154.2564$

$K_c=\frac{282.5761\times 10^{36}}{154.2564\times 10^{-42}}$

$K_c=\frac{282.5761}{154.2564}\times10^{36 + 42}$

$K_c\approx1.83\times 10^{78}$

Answer:

$1.83\times 10^{78}$