Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 15: Problem 24

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 \(K_{p}\). (b) Does the equilibrium mixture contain mostly \(\mathrm{H}_{2}\) and \(\mathrm{S}_{2}\) or mostly \(\mathrm{H}_{2} \mathrm{~S} ?\) (c) Calculate the values of \(\mathrm{K}_{c}\) and \(K_{p}\) if you rewrote the balanced chemical equation with \(1 \mathrm{~mol}\) of \(\mathrm{H}_{2}(g)\) instead of \(2 \mathrm{~mol}\).

Short Answer

Expert verified
In summary, for the given equilibrium reaction, we find: a) The value of \(K_p\) is approximately \(5.02 \times 10^{5}\). b) The equilibrium mixture contains mostly \(\mathrm{H}_{2}\mathrm{S}\), due to the large value of \(K_c\). c) For the modified balanced chemical equation, the new values of \(K'_c\) and \(K'_p\) are approximately \(3.28 \times 10^3\) and \(3.55 \times 10^2\), respectively.

Step by step solution

01

Recall the relationship between K_c and K_p

We can calculate \(K_p\) using the following equation: \(K_p = K_c(RT)^{\Delta{ n}}\) where: - \(K_p\) is the equilibrium constant in terms of pressure - \(K_c\) is the equilibrium constant in terms of concentration, which is given as \(1.08 \times 10^{7}\) - \(R\) is the ideal gas constant, which is \(8.314~\mathrm{J} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}\) - \(T\) is the temperature in Kelvin - \(\Delta{n}\) is the change in the number of moles of gas in the reaction
02

Find the change in the number of moles of gas, Δn

Calculate the change in the number of moles of gas in the reaction: \(\Delta n = (\text{moles of product}) - (\text{moles of reactant})\) For the given reaction: \(\Delta n = (2~\text{moles of}\, \mathrm{H}_{2} \mathrm{~S}(g)) - (2~\text{moles of}\, \mathrm{H}_{2}(g) + 1~\text{mole of}\, \mathrm{S}_{2}(g))\) \(\Delta n = 2 - 3 = -1\)
03

Convert the temperature to Kelvin

We need to convert the given temperature, which is \(700^{\circ}\mathrm{C}\), to Kelvin: \(T(K) = 700 + 273.15\) \(T(K) = 973.15~\mathrm{K}\)
04

Calculate K_p

Replace the constants and variables values in the equation: \(K_p = (1.08 \times 10^{7})(8.314)(973.15)^{-1}\) Now, solve for \(K_p\): \(K_p \approx 5.02 \times 10^{5}\) So, the value of \(K_p\) is approximately \(5.02 \times 10^{5}\). #b) Composition of the Equilibrium Mixture#
05

Analyze the value of K_c

The given value of \(K_c\) is \(1.08 \times 10^7\), which is a large number. A large value for \(K_c\) indicates that at equilibrium, the concentration of products is higher than the concentration of reactants. In this case, the equilibrium mixture contains mostly \(\mathrm{H}_{2}\mathrm{S}\). #c) Calculating new K_c and K_p values for the modified balanced chemical equation#
06

Write the modified balanced chemical equation

The modified balanced chemical equation contains \(1 \mathrm{~mol}\) of \(\mathrm{H}_{2}(g)\) instead of \(2 \mathrm{~mol}\): \(\mathrm{H}_{2}(g)+\frac{1}{2}\mathrm{S}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{~S}(g)\)
07

Calculate the new K_c value

The new balanced chemical equation is 1/2 the original equation. Thus, the new \(K'_c\) value will be the square root of the original \(K_c\) value: \(K'_c = \sqrt{1.08 \times 10^7}\) \(K'_c \approx 3.28 \times 10^3\) So, the new value of \(K'_c\) is approximately \(3.28 \times 10^3\).
08

Calculate the new K_p value

The change in the number of moles of gas, \(\Delta n'\), for the modified balanced chemical equation is: \(\Delta n' = 1 - 1.5 = -0.5\) Using the new \(K'_c\) value and the modified \(\Delta n'\) value, calculate the new \(K'_p\) value: \(K'_p = K'_c(RT)^{\Delta{ n'}}\) \(K'_p = (3.28 \times 10^{3})(8.314)(973.15)^{-0.5}\) Solve for \(K'_p\): \(K'_p \approx 3.55 \times 10^2\) So, the new value of \(K'_p\) for the modified balanced chemical equation is approximately \(3.55 \times 10^2\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A friend says that the faster the reaction, the larger the equilibrium constant. Is your friend correct? Why or why not? [Sections \(15 \cdot 1\) and 15.2\(]\)

A \(0.831-\mathrm{g}\) sample of \(\mathrm{SO}_{3}\) is placed in a 1.00 - \(\mathrm{L}\) container and heated to \(1100 \mathrm{~K}\). The \(\mathrm{SO}_{3}\) decomposes to \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) : $$2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)$$ At equilibrium the total pressure in the container is \(1.300 \mathrm{~atm}\). Find the values of \(K_{p}\) and \(K_{c}\) for this reaction at \(1100 \mathrm{~K}\).

At \(700 \mathrm{~K}\) the equilibrium constant for the reaction $$\mathrm{CCl}_{4}(g) \rightleftharpoons \mathrm{C}(s)+2 \mathrm{Cl}_{2}(g)$$ is \(K_{p}=0.76 .\) A flask is charged with 2.00 atm of \(\mathrm{CCl}_{4}\), which then reaches equilibrium at \(700 \mathrm{~K}\). (a) What fraction of the \(\mathrm{CCl}_{4}\) is converted into \(\mathrm{C}\) and \(\mathrm{Cl}_{2} ?\) (b) What are the partial pressures of \(\mathrm{CCl}_{4}\) and \(\mathrm{Cl}_{2}\) at equilibrium?

Consider the reaction $$\mathrm{CaSO}_{4}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q)$$ At \(25^{\circ} \mathrm{C}\) the equilibrium constant is \(K_{c}=2.4 \times 10^{-5}\) for this reaction. (a) If excess \(\mathrm{CaSO}_{4}(s)\) is mixed with water at \(25^{\circ} \mathrm{C}\) to produce a saturated solution of \(\mathrm{CaSO}_{4},\) what are the equilibrium concentrations of \(\mathrm{Ca}^{2+}\) and \(\mathrm{SO}_{4}^{2-} ?(\mathbf{b})\) If the resulting solution has a volume of \(1.4 \mathrm{~L},\) what is the minimum mass of \(\mathrm{CaSO}_{4}(s)\) needed to achieve equilibrium?

At \(100^{\circ} \mathrm{C}\) the equilibrium constant for the reaction \(\mathrm{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g)\) has the value \(K_{c}=\) \(2.19 \times 10^{-10} .\) Are the following mixtures of \(\mathrm{COCl}_{2}, \mathrm{CO}\) and \(\mathrm{Cl}_{2}\) at \(100^{\circ} \mathrm{C}\) at equilibrium? If not, indicate the direction that the reaction must proceed to achieve equilibrium. (a) \(\quad\left[\mathrm{COCl}_{2}\right]=2.00 \times 10^{-3} \mathrm{M}, \quad[\mathrm{CO}]=3.3 \times 10^{-6} \mathrm{M},\) \(\left[\mathrm{Cl}_{2}\right]=6.62 \times 10^{-6} \mathrm{M} ;\) (b) \(\left[\mathrm{COCl}_{2}\right]=4.50 \times 10^{-2} \mathrm{M}\) \([\mathrm{CO}]=1.1 \times 10^{-7} \mathrm{M},\left[\mathrm{Cl}_{2}\right]=2.25 \times 10^{-6} \mathrm{M} ;(\mathrm{c})\left[\mathrm{COCl}_{2}\right]=\) \(0.0100 M,[\mathrm{CO}]=\left[\mathrm{Cl}_{2}\right]=1.48 \times 10^{-6} \mathrm{M}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Making Measurements

Read Explanation

Chemical Analysis

Read Explanation

Organic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Nuclear Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free