Consider the following equilibrium: $2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{~S}(g) \quad K_{c}=1.08 \times 10^{7}\( at \)700^{\circ} \mathrm{C}$ (a) Calculate \(K_{p \cdot}\) (b) Does the equilibrium mixture contain mostly \(\mathrm{H}_{2}\) and \(\mathrm{S}_{2}\) or mostly $\mathrm{H}_{2} \mathrm{~S} ?(\mathbf{c})\( Calculatethevalue of \)K_{c}$ if you rewrote the equation $\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{~S}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{~S}(g)$

Short Answer

Expert verified
(a) The value of \(K_p\) at \(700^{\circ} \mathrm{C}\) is \(2.68\times10^8\). (b) The equilibrium mixture mostly contains H2S. (c) The new value of \(K_c\) for the rewritten equation is \(3.286 \times 10^3\).

Step by step solution

01

Convert Kc to Kp

To convert the equilibrium constant Kc to Kp, we use the following equation: \(K_p = K_c(RT)^{\Delta n}\), where R is the ideal gas constant (0.0821 atm L/mol K), T is the temperature in kelvin, and Δn is the change in moles of gas (products - reactants). First, we need to convert the temperature from Celsius to Kelvin: \(T = 700 + 273.15 = 973.15K\) Next, we calculate Δn: \(\Delta n = (2 - (2 + 1)) = -1\) Now, we can calculate Kp: \(K_p = K_c (RT)^{\Delta n} = (1.08 \times 10^7)((0.0821)(973.15))^{-1}\)
02

Calculate Kp

By plugging the values into the Kp formula, we get: \(K_p = (1.08 \times 10^7)(0.0799)^{-1} \Rightarrow K_p = 2.68 \times 10^8\) So, the value of Kp at 700°C is \(2.68\times10^8\).
03

Determine the composition of the equilibrium mixture

To determine the composition of the equilibrium mixture, we can compare the Kc value with the stoichiometry of the reaction. A Kc value significantly larger than 1 indicates that the reaction favors the products, while a Kc value significantly smaller than 1 indicates that the reaction favors the reactants. Since the given Kc value is very large at \(1.08 \times 10^7\), we can conclude that the equilibrium mixture will mostly contain H2S.
04

Calculate the new Kc for the rewritten equation

When the reaction is rewritten as \(\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{~S}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{~S}(g)\), the stoichiometric coefficients are halved relative to the original reaction. To find the new Kc value, we must square root the original Kc. New Kc = \(\sqrt{1.08 \times 10^7} \Rightarrow\) New Kc = \(3.286 \times 10^3\) The new value of Kc for the rewritten equation is \(3.286 \times 10^3\).

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Most popular questions from this chapter

Bromine and hydrogen react in the gas phase to form hydrogen bromide: \(\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g) .\) The reaction enthalpy is \(\Delta H^{\circ}=-6 \mathrm{~kJ} .\) (a) To increase the equilibrium yield of hydrogen bromide would you use high or low temperature? (b) Could you increase the equilibrium yield of hydrogen bromide by controlling the pressure of this reaction? If so, would high or low pressure favor formation of \(\mathrm{HBr}(g) ?\)

At \(800 \mathrm{~K},\) the equilibrium constant for $\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)\( is \)K_{c}=3.1 \times 10^{-5}$. If an equilibrium mixture in a 5.00-L vessel contains \(30.5 \mathrm{mg}\) of \(\mathrm{I}(g)\), how many grams of \(\mathrm{I}_{2}\) are in the mixture?

The reaction of an organic acid with an alcohol, in organic solvent, to produce an ester and water is commonly done in the pharmaceutical industry. This reaction is catalyzed by strong acid (usually $\left.\mathrm{H}_{2} \mathrm{SO}_{4}\right)$. A simple example is the reaction of acetic acid with ethyl alcohol to produce ethyl acetate and water: $$\begin{aligned} \mathrm{CH}_{3} \mathrm{COOH}(\mathrm{solv})+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(\mathrm{solv}) & \mathrm{CH}_{3} \mathrm{COOCH}_{2} \mathrm{CH}_{3}(\text { solv })+\mathrm{H}_{2} \mathrm{O}(\text { solv }) \end{aligned}$$ where "(solv)" indicates that all reactants and products are in solution but not an aqueous solution. The equilibrium constant for this reaction at \(55^{\circ} \mathrm{C}\) is 6.68 . A pharmaceutical chemist makes up $15.0 \mathrm{~L}\( of a solution that is initially \)0.275 \mathrm{M}$ in acetic acid and \(3.85 \mathrm{M}\) in ethanol. At equilibrium, how many grams of ethyl acetate are formed?

Consider the following equilibrium between oxides of nitrogen $$3 \mathrm{NO}(g) \rightleftharpoons \mathrm{NO}_{2}(g)+\mathrm{N}_{2} \mathrm{O}(g)$$ (a) Use data in Appendix C to calculate \(\Delta H^{\circ}\) for this reaction. (b) Will the equilibrium constant for the reaction increase or decrease with increasing temperature? (c) At constant temperature, would a change in the volume of the container affect the fraction of products in the equilibrium mixture?

(a) If \(Q_{c}>K_{c}\), how must the reaction proceed to reach equilibrium? (b) At the start of a certain reaction, only reactants are present; no products have been formed. What is the value of \(Q_{c}\) at this point in the reaction?

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