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Chapter 15: Problem 26

Consider the following equilibrium, for which \(K_{p}=7.62\) at $480^{\circ} \mathrm{C}:$ $$2 \mathrm{Cl}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 4 \mathrm{HCl}(g)+\mathrm{O}_{2}(g)$$ (a) What is the value of \(K_{p}\) for the reaction $4 \mathrm{HCl}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{Cl}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) ?$ (b) What is the value of \(K_{p}\) for the reaction $\mathrm{Cl}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HCl}(g)+\frac{1}{2} \mathrm{O}_{2}(g) ?$ (c) What is the value of \(K_{c}\) for the reaction in part (b)?

Short Answer

Expert verified
(a) The \(K_p\) for the reverse reaction is 0.1312. (b) The new \(K_p\) after adjusting coefficients is 2.760. (c) The \(K_c\) value for the reaction in part (b) is 0.284.

Step by step solution

01

(a) \(K_p\) for the reverse reaction

The \(K_p\) for the original reaction is given as 7.62. Since the reverse reaction is the inverse of the original, the \(K_p\) for the reverse reaction is the reciprocal of the original \(K_p\). Therefore: \(K_p' = \frac{1}{K_p} = \frac{1}{7.62}\)
02

(a) Reverse reaction \(K_p\) answer

Calculating the inverse of 7.62, we get \(K_p' = 0.1312\)
03

(b) Coefficient change impact on \(K_p\)

Given the reaction: $$\mathrm{Cl}_{2}(g)+\mathrm{H}_{2}\mathrm{O}(g) \rightleftharpoons 2\mathrm{HCl}(g)+\frac{1}{2}\mathrm{O}_{2}(g)$$ The coefficients have been divided by 2 compared to the original reaction. Therefore, we raise the original \(K_p\) to power of the factor (1/2) to get the new \(K_p\) value: \(K_p'' = (K_p)^{\frac{1}{2}} = (7.62)^{\frac{1}{2}}\)
04

(b) New \(K_p\) after adjusting coefficients

Calculating the square root of 7.62, we get \(K_p'' = 2.760\)
05

(c) Converting \(K_p\) to \(K_c\)

To convert our \(K_p\) value for part (b) to a \(K_c\) value, we use the formula \(K_p = K_c(RT)^{\Delta n}\). We need to calculate the \(\Delta n\) value: $$\Delta n = n_{products} - n_{reactants} = (2 + \frac{1}{2}) - (1 + 1) = \frac{1}{2}$$ We are given that T = 480℃, and we should convert it to Kelvin: T = 480 + 273.15 = 753.15 K. The gas constant, R, has a value of 0.0821 L atm K⁻¹ mol⁻¹. Now, we can solve for \(K_c\): $$K_c = \frac {K_p}{(RT)^{\Delta n}} = \frac {2.760}{(0.0821 \times 753.15)^{\frac{1}{2}}}$$
06

(c) Calculating the \(K_c\) value

Calculating the \(K_c\) value, we find $$K_c = 2.760 \times \frac{1}{(0.0821 \times 753.15)^{\frac{1}{2}}} = 0.284$$ So, the \(K_c\) value for the reaction in part (b) is 0.284.

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Most popular questions from this chapter

At \(1285^{\circ} \mathrm{C}\), the equilibrium constant for the reaction \(\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{Br}(g)\) is $K_{c}=1.04 \times 10^{-3} .\( A \)1.00-\mathrm{L}$ vessel containing an equilibrium mixture of the gases has \(1.50 \mathrm{~g}\) \(\mathrm{Br}_{2}(g)\) in it. What is the mass of \(\mathrm{Br}(g)\) in the vessel?

The equilibrium $2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{NOCl}(g)\( is established at \)550 \mathrm{~K}$. An equilibrium mixture of the three gases has partial pressures of $10.13 \mathrm{kPa}, 20.27 \mathrm{kPa}\(, and \)35.46 \mathrm{kPa}\( for \)\mathrm{NO}, \mathrm{Cl}_{2},$ and \(\mathrm{NOCl}\), respectively.(a) Calculate \(K_{p}\) for this reaction at \(500.0 \mathrm{~K}\). (b) If the vessel has a volume of \(5.00 \mathrm{~L},\) calculate \(K_{c}\) at this temperature.

Consider the reaction $\mathrm{IO}_{4}^{-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\( \)\mathrm{H}_{4} \mathrm{IO}_{6}^{-}(a q) ; K_{c}=3.5 \times 10^{-2} .\( If you start with \)25.0 \mathrm{~mL}\( of a \)0.905 \mathrm{M}\( solution of \)\mathrm{NaIO}_{4}$, and then dilute it with water to \(500.0 \mathrm{~mL},\) what is the concentration of $\mathrm{H}_{4} \mathrm{IO}_{6}^{-}$ at equilibrium?

Which of the following statements are true and which are false? (a) For the reaction $2 \mathrm{~A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{A}_{2} \mathrm{~B}(g) K_{c}$ and \(K_{p}\) are numerically the same. (b) It is possible to distinguish \(K_{c}\) from \(K_{p}\) by comparing the units used to express the equilibrium constant. \((\mathbf{c})\) For the equilibrium in (a), the value of \(K_{c}\) increases with increasing pressure.

Assume that the equilibrium constant for the dissociation of molecular bromine, \(\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{Br}(g)\), at 800 \(\mathrm{K}\) is \(K_{c}=5.4 \times 10^{-3}\). (a) Which species predominates at equilibrium, \(\mathrm{Br}_{2}\) or Br, assuming that the concentration of \(\mathrm{Br}_{2}\) is larger than $5.4 \times 10^{-3} \mathrm{~mol} / \mathrm{L} ?$ (b) Assuming both forward and reverse reactions are elementary processes, which reaction has the larger numeric value of the rate constant, the forward or the reverse reaction?
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