Question

help me by showing the work please

Given the following information, Kc 2.0 x 10-38 Kc 2.4 x 101 2 NO(g) O2(g) 2 NO2 find the value of Ke for the following reaction: NO(g) + 03(g) NO2(g) + O2(g) A. 5.6 x 10-17 B. 3.5 x 1025 C. 9.0 x 1040 D. 1.1 x 10-4 practice Given the following information at 25°C, 2 NOCIg)2 NO(g) + Cl2(g) Kc 4.0 x 102 moV/L find the value of K for this reaction at 25°C A. 0.69 atm B. 0.98 atm C. 24 atm D. 1.6 x 10-3 atm

Answer 1

1.

The reaction for which we have to find Kc is

$NO_{(g)} + O_3_{(g)} \rightleftharpoons NO_2_{(g)} + O_2_{(g)}--------(0)$

For a general reaction

aA + bB cC + dD

the expression of Kc is

$\frac{[C]^c \times [D]^d}{[A]^a \times [B]^b}$

Hence, for our reaction Kc should be

$K_c = \frac{[NO_2] \times [O_2]}{[NO]\times [O_3]}$

Now, from the given reactions with Kc value

02g)20sg)

(03)2 0213 cl = 2.0 × 10-38

Similarly for $2NO_{(g)} + O_2 _{(g)} \rightleftharpoons 2NO_2_{(g)}-------------(2)$

[N0212 = 2.4 × 1013

Now looking at the expression of Kc for the required reaction

$K_c = \frac{[NO_2] \times [O_2]}{[NO]\times [O_3]}$

We have to find $K_c$ in terms of .

Kcl and Kc2

Hence, we will first write the equation(0) in terms of equation (1) and (2).

O3 is in the reactant side in equation (0) and in the product side in equation (1)

Hence, we must reverse the equation in equation (1)

$2O_3_{(g)} \rightleftharpoons 3O_2_{(g)}-------------(3)$

Hence, the new Kc1 of the reverse reaction is

$K_{c1}' = \frac{[O_2]^3}{[O_3]^2} = \frac{1}{K_{c1}}$

There is only one mole of O3 in equation (0) in the product side, hence we must divide equation (3) by 2.

[20)302)] 3(g)

Hence, the new Kc1' is

23/2 Kc1

Now our required equation (0) has 1 mole of NO on reactant side.

Hence, we must divide equation (2) by 2 to get 1 mole NO

02g)-(5)

Hence, Kc for equation (5).

$K_{c2}' = \frac{[NO_2]}{[NO]\times [O_2]^{(1/2)}} = \sqrt{(K_{c2})}$

Now, adding equation (4) and (5) we get the equation (0).

$O_3_{(g)} \rightleftharpoons \frac{3}{2} O_2{(g)} --------(4) \\ + NO_{(g)} + \frac{1}{2}O_2_{(g)} \rightleftharpoons NO_2_{(g)}-----(5) \\ --------------------\\ NO_{(g)} + O_3_{(g)} \rightleftharpoons NO_2_{(g)} + O_2_{(g)}$

Hence, the Kc of equation (0) will be the product of equation (4) and equation (5) as shown below

$K_c = \sqrt{(1/K_{c1})} \times \sqrt{K_{c2}} =\sqrt{\left (\frac{[O_2]^3} {[O_3]^2} \right )} \times \sqrt{\left ( \frac{[NO_2]^2}{[NO]^2 \times[O_2]} \right )} \\ \Rightarrow K_c = \frac{[O_2]^{(3/2)} \times [NO_2]}{[O_3] \times [NO] \times [O_2]^{(1/2)}} = \frac{[NO_2] \times [O_2]}{[NO]\times [O_3]}$

Which is exactly what it should be.

Hence,

Given that

$K_{c1} = 2.0 \times 10^{-38} \\ \ and \ K_{c2} = 2.4 \times 10^{13} \\$

The required Kc is

$K_c = \sqrt{(1/K_{c1})} \times \sqrt{K_{c2}} = \sqrt{\frac{1}{2.0 \times 10^{-38}}} \times \sqrt{2.4 \times 10^{13}} =\sqrt{\frac{2.4}{2.0} \times 10^{51}} = 3.46 \times 10^{25} \approx 3.5 \times 10^{25}$

hence, the answer is B $\approx 3.5 \times 10^{25}$ .

2.

The given equilibrium is

$2NOCl_{(g)} \rightleftharpoons 2NO_{(g)} + Cl_2_{(g)}$

Given that

$T = 25 ^\circ C = 25 + 273.15 \ K = 298.15 \ K$

$K_c = 4.0 \times 10^{-2} \ mol/L$

we have to find out Kp, which is related to Kc by the following equation

$K_p = K_c(RT)^{\Delta n_{g}}$

${\Delta n_{g}}$ is the number of moles of gaseous products minus number of moles of gaseous reactant.

Hence, ${\Delta n_{g}} = 2 \ mol (NO_{(g)}) + 1 \ mol (Cl_2_{(g)}) - 2 \ mol \ (NOCl_{(g)}) = 1 \ mol$

Gas constant R =

Hence, putting everything in

$K_p = K_c(RT)^{\Delta n_{g}} \\ \Rightarrow K_p = 4.0 \times 10^{-2} \ mol \ L^{-1} (0.082 \ L \ atm \ K^{-1} \ mol^{-1} \times 298.15 \ K)^1 \\ \Rightarrow K_p = 0.9779 \ atm \approx 0.98 \ atm$

hence, the correct option is B.0.98 atm