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Chapter 12: Problem 24

For the reaction $$\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g) $$.\(K_{\mathrm{p}}=3.5 \times 10^{4}\) at \(1495 \mathrm{K} .\) What is the value of \(K_{\mathrm{p}}\) for the following reactions at \(1495 \mathrm{K} ?\) a. \(\operatorname{HBr}(g) \rightleftharpoons \frac{1}{2} \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{Br}_{2}(g)\). b. \(2 \mathrm{HBr}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g)\) c. \(\frac{1}{2} \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{Br}_{2}(g) \rightleftharpoons \mathrm{HBr}(g)\).

Short Answer

Expert verified
For the three reactions at 1495 K, the values of Kp are: a. Kp = 0.00535 b. Kp = 2.857 × 10^(-5) c. Kp = 187.082

Step by step solution

01

Find the reaction relation with the given equation

Observe that this reaction is the reverse of the given reaction and divided by 2.
02

Calculate the equilibrium constant for the reverse reaction

Since the reaction is reversed, we need to find the reciprocal of the equilibrium constant of the given reaction Kp_reverse = 1/Kp = 1/(3.5 × 10^4) = 2.857 × 10^(-5).
03

Calculate the equilibrium constant for the new reaction

Next, raise the Kp_reverse to the power of the fraction by which the reaction has been divided: Kp_new = (Kp_reverse)^(1/2) = (2.857 × 10^(-5))^(1/2) = 0.00535. Answer: Kp = 0.00535 for reaction a at 1495 K. b. 2 HBr(g) <=> H2(g) + Br2(g)
04

Find the reaction relation with the given equation

This reaction corresponds to the reverse of the given reaction.
05

Calculate the equilibrium constant for the reversed reaction

Since the reaction is reversed, we need to find the reciprocal of the given equilibrium constant: Kp_reverse = 1/Kp = 1/(3.5 × 10^4) = 2.857 × 10^(-5). Answer: Kp = 2.857 × 10^(-5) for reaction b at 1495 K. c. (1/2) H2(g) + (1/2) Br2(g) <=> HBr(g)
06

Find the reaction relation with the given equation

Observe that this reaction is the given reaction divided by 2.
07

Calculate the equilibrium constant for the new reaction

Raise the given equilibrium constant Kp to the power of the fraction by which the reaction has been divided: Kp_new = (Kp)^(1/2) = (3.5 × 10^4)^(1/2) = 187.082. Answer: Kp = 187.082 for reaction c at 1495 K.

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Most popular questions from this chapter

The equilibrium constant is 0.0900 at \(25^{\circ} \mathrm{C}\) for the reaction $$ \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Cl}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HOCl}(g)$$.For which of the following sets of conditions is the system at equilibrium? For those that are not at equilibrium, in which direction will the system shift? a. A 1.0 -L flask contains 1.0 mole of HOCI, 0.10 mole of \(\mathrm{Cl}_{2} \mathrm{O},\) and 0.10 mole of \(\mathrm{H}_{2} \mathrm{O}\) b. A \(2.0-\) L flask contains 0.084 mole of HOCI, 0.080 mole of \(\mathrm{Cl}_{2} \mathrm{O},\) and 0.98 mole of \(\mathrm{H}_{2} \mathrm{O}\) c. A 3.0 -L flask contains 0.25 mole of HOCI, 0.0010 mole of \(\mathrm{Cl}_{2} \mathrm{O},\) and 0.56 mole of \(\mathrm{H}_{2} \mathrm{O}\).

Consider the reaction$$\mathrm{Fe}^{3+}(a q)+\mathrm{SCN}^{-}(a q) \rightleftharpoons \mathrm{FeSCN}^{2+}(a q)$$.How will the equilibrium position shift if a. water is added, doubling the volume? b. \(\mathrm{AgNO}_{3}(a q)\) is added? (AgSCN is insoluble.) c. \(\mathrm{NaOH}(a q)\) is added? \(\left[\mathrm{Fe}(\mathrm{OH})_{3} \text { is insoluble. }\right]\) d. \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}(a q)\) is added? c. \(\mathrm{NaOH}(a q)\) is added?is insoluble.] d. \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}(a q)\) is added?

At \(35^{\circ} \mathrm{C}, K=1.6 \times 10^{-5}\) for the reaction $$2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g)$$.If 2.0 moles of \(\mathrm{NO}\) and 1.0 mole of \(\mathrm{Cl}_{2}\) are placed into a \(1.0-\mathrm{L}\) flask, calculate the equilibrium concentrations of all species.

Predict the shift in the equilibrium position that will occur for each of the following reactions when the volume of the reaction container is increased. a. \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) b. \(\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)\) c. \(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \rightleftharpoons 2 \mathrm{HF}(g)\) d. \(\operatorname{COCl}_{2}(g) \rightleftharpoons \operatorname{CO}(g)+\mathrm{Cl}_{2}(g)\) e. \(\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)\)

In which direction will the position of the equilibrium.$$2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g).$$,be shifted for each of the following changes? a. \(\mathrm{H}_{2}(g)\) is added. b. \(\mathrm{I}_{2}(g)\) is removed. c. \(\mathrm{HI}(g)\) is removed. d. In a rigid reaction container, some \(\operatorname{Ar}(g)\) is added. e. The volume of the container is doubled. f. The temperature is decreased (the reaction is exothermic).
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