Question

1. the equilibrium constant, $k_c$, is given for one of the reactions below. what is the value of the missing equilibrium constant, $k_c$?.$h_2(g) + br_2(g) \

ightleftharpoons 2 hbr(g)$ $k_c = 3.8 \times 10^4

$$4 hbr(g) \ ightleftharpoons 2 h_2(g) + 2 br_2(g)$ $k_c =?$a. $k_c = 1.9 \\times 10^4$b. $k_c = 5.1 \\times 10^{-3}$c. $k_c = 2.6 \\times 10^{-5}$d. $k_c = 6.9 \\times 10^{-10}$e. $k_c = 1.6 \\times 10^3$18. at $35^\\circ c$, the equilibrium constant for the reaction below is $k_c = 1.6 \\times 10^{-5}$.$2 nocl(g) \ ightleftharpoons 2 no(g) + cl_2(g)$calculate $no$ at equilibrium if the mixture contains $cl_2 = 1.2 \\times 10^{-2}\\ m$ and $nocl = 2.8 \\times 10^{-1}\\ m$ at equilibrium.a. $1.0 \\times 10^{-4}\\ m$b. $1.0 \\times 10^{-2}\\ m$c. $2.8 \\times 10^{-1}\\ m$d. $2.4 \\times 10^{-2}\\ m$e. $1.6 \\times 10^{-3}\\ m$19. given the following equilibria:$pb^{2+}(aq) + 2 br^-(aq) \ ightleftharpoons pbbr_2(s)$ $k_c = 1.51 \\times 10^5$$

pb^{2+}(aq) + 2 oh^-(aq) \
ightleftharpoons pb(oh)_2(s)$ $k_c = 7.14 \times 10^{14}$determine the equilibrium constant, $k_c$, for the reaction below.$pb(oh)_2(s) + 2 br^-(aq) \
ightleftharpoons pbbr_2(s) + 2 oh^-(aq)$a. $k_c = 1.1 \times 10^{20}$b. $k_c = 4.7 \times 10^9$c. $k_c = 6.6 \times 10^{-6}$d. $k_c = 2.1 \times 10^{-10}$e. $k_c = 9.2 \times 10^{-21}$

Explanation:

Question 17

Step1: Relate reactions (reverse + scale)

The target reaction is 2×reverse of given reaction. For reversed reaction, $K'_c = \frac{1}{K_c}$. Scaling by 2 gives $K = (K'_c)^2$.

Step2: Calculate reversed $K_c$

$K'_c = \frac{1}{3.8 \times 10^4} \approx 2.63 \times 10^{-5}$

Step3: Scale to target reaction

$K_c = (2.63 \times 10^{-5})^2 \approx 6.9 \times 10^{-10}$

Step1: Write $K_c$ expression

For $2\text{NOCl}(g)
ightleftharpoons 2\text{NO}(g) + \text{Cl}_2(g)$,
$K_c = \frac{[\text{NO}]^2 [\text{Cl}_2]}{[\text{NOCl}]^2}$

Step2: Rearrange for $[\text{NO}]^2$

$[\text{NO}]^2 = K_c \times \frac{[\text{NOCl}]^2}{[\text{Cl}_2]}$

Step3: Substitute known values

$[\text{NO}]^2 = 1.6 \times 10^{-5} \times \frac{(2.8 \times 10^{-1})^2}{1.2 \times 10^{-2}}$
$= 1.6 \times 10^{-5} \times \frac{7.84 \times 10^{-2}}{1.2 \times 10^{-2}} \approx 1.045 \times 10^{-4}$

Step4: Solve for $[\text{NO}]$

$[\text{NO}] = \sqrt{1.045 \times 10^{-4}} \approx 1.0 \times 10^{-2}\ \text{M}$

Step1: Combine given equilibria

Reverse the second reaction: $\text{Pb(OH)}_2(s)
ightleftharpoons \text{Pb}^{2+}(aq) + 2\text{OH}^-(aq)$, $K'_c = \frac{1}{7.14 \times 10^{14}}$
Add to first reaction: $\text{Pb}^{2+}(aq) + 2\text{Br}^-(aq)
ightleftharpoons \text{PbBr}_2(s)$, $K_1 = 1.51 \times 10^5$

Step2: Calculate combined $K_c$

$K_c = K_1 \times K'_c = \frac{1.51 \times 10^5}{7.14 \times 10^{14}} \approx 2.1 \times 10^{-10}$

Answer:

D. $K_c = 6.9 \times 10^{-10}$

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Question 18