/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 Write the chemical equations cor... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Write the chemical equations corresponding to the following equilibrium expressions. (a) \(K=\frac{\left[\mathrm{H}_{2} \mathrm{CO}_{3}\right]}{\left[\mathrm{CO}_{2}\right]\left[\mathrm{H}_{2} \mathrm{O}\right]}\) (b) \(K=\frac{\left[\mathrm{P}_{4}\right]\left[\mathrm{O}_{2}\right]^{5}}{\left[\mathrm{P}_{4} \mathrm{O}_{10}\right]}\) (c) \(K=\frac{\left[\mathrm{F}_{2}\right]^{3}\left[\mathrm{PH}_{3}\right]}{[\mathrm{HF}]^{3}\left[\mathrm{PF}_{3}\right]}\)

Short Answer

Expert verified
(a) \( \mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{H}_{2}\mathrm{CO}_{3} \); (b) \( \mathrm{P}_{4} + 5\mathrm{O}_{2} \rightleftharpoons \mathrm{P}_{4}\mathrm{O}_{10} \); (c) \( 3\mathrm{HF} + \mathrm{PF}_{3} \rightleftharpoons 3\mathrm{F}_{2} + \mathrm{PH}_{3} \).

Step by step solution

01

Understand the Equilibrium Constant Expression

The equilibrium constant expression is the ratio of the concentrations of the products to the reactants, raised to their respective coefficients in the balanced chemical equation. Let's analyze each of the given equilibrium constant expressions to figure out the chemical equations.
02

Derive Chemical Equation for Part (a)

The given expression is \( K = \frac{[\mathrm{H}_{2}\mathrm{CO}_{3}]}{[\mathrm{CO}_{2}][\mathrm{H}_{2}\mathrm{O}]} \). The numerator is the product \([\mathrm{H}_{2}\mathrm{CO}_{3}]\) and the denominator has the reactants \([\mathrm{CO}_{2}]\) and \([\mathrm{H}_{2}\mathrm{O}]\). Thus, the chemical equation is: \[ \mathrm{CO}_{2} (aq) + \mathrm{H}_{2}\mathrm{O} (l) \rightleftharpoons \mathrm{H}_{2}\mathrm{CO}_{3} (aq) \]
03

Derive Chemical Equation for Part (b)

The given expression is \( K = \frac{[\mathrm{P}_{4}][\mathrm{O}_{2}]^{5}}{[\mathrm{P}_{4}\mathrm{O}_{10}]} \). This indicates that \( \mathrm{P}_{4} \) and \( \mathrm{O}_{2} \) are reactants, and \( \mathrm{P}_{4}\mathrm{O}_{10} \) is the product. The balanced chemical equation is: \[ \mathrm{P}_{4} (s) + 5 \mathrm{O}_{2} (g) \rightleftharpoons \mathrm{P}_{4}\mathrm{O}_{10} (s) \]
04

Derive Chemical Equation for Part (c)

The given expression is \( K = \frac{[\mathrm{F}_{2}]^{3}[\mathrm{PH}_{3}]}{[\mathrm{HF}]^{3}[\mathrm{PF}_{3}]} \). This suggests that \( \mathrm{HF} \) and \( \mathrm{PF}_{3} \) are reactants, while \( \mathrm{F}_{2} \) and \( \mathrm{PH}_{3} \) are products. The balanced chemical equation should be: \[ 3 \mathrm{HF} (g) + \mathrm{PF}_{3} (g) \rightleftharpoons 3 \mathrm{F}_{2} (g) + \mathrm{PH}_{3} (g) \]

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.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Constant
The equilibrium constant, represented by the symbol \(K\), is a crucial concept in understanding chemical equilibrium. It charts the balance between reactants and products in a reversible chemical reaction at a specific temperature. The value of \(K\) provides insight into the ratio of concentrations of products to reactants when the reaction has reached equilibrium.

In mathematical terms, the equilibrium constant for a general reaction \(aA + bB \rightleftharpoons cC + dD\) is given by:\[ K = \frac{[C]^c[D]^d}{[A]^a[B]^b} \]Here, the square brackets denote the concentrations of the substances at equilibrium, and the letters \(a\), \(b\), \(c\), and \(d\) are the stoichiometric coefficients from the balanced equation.

Key points about \(K\):
  • If \(K > 1\), the equilibrium position favors products.
  • If \(K < 1\), the equilibrium position favors reactants.
  • \(K\) does not change unless the temperature changes.
Chemical Equation
Chemical equations are symbolic representations of chemical reactions. They consist of reactants, products, and an arrow that indicates the direction of the reaction.

In a balanced chemical equation, the number of atoms for each element is the same on both sides, ensuring that mass is conserved according to the law of conservation of mass. For example, the equation for the decomposition of water into hydrogen and oxygen gases is:\[ 2\mathrm{H}_2\mathrm{O} \rightarrow 2\mathrm{H}_2 + \mathrm{O}_2 \]This equation shows that two water molecules break down into two hydrogen molecules and one oxygen molecule.

In reversible reactions, indicated by a double arrow (\(\rightleftharpoons\)), the reaction can proceed in both directions. For instance, the conversion of carbon dioxide and water into carbonic acid, as described by the equation:\[ \mathrm{CO}_2 (aq) + \mathrm{H}_2\mathrm{O} (l) \rightleftharpoons \mathrm{H}_2\mathrm{CO}_3 (aq) \]
Reactants and Products
The terms "reactants" and "products" refer to the starting materials and the substances generated in a chemical reaction, respectively.

Reactants are the substances initially present before the chemical change. For example, in the reaction \(\mathrm{P}_4 (s) + 5 \mathrm{O}_2 (g) \rightleftharpoons \mathrm{P}_4 \mathrm{O}_{10} (s)\), phosphorus (\(\mathrm{P}_4\)) and oxygen (\(\mathrm{O}_2\)) are the reactants.

Products are the new substances formed as a result. In the same reaction, \(\mathrm{P}_4 \mathrm{O}_{10}\) is the product formed.

Understanding the role of reactants and products helps us write equilibrium constant expressions correctly, ensuring that we accurately describe the chemistry of the system.

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

As we shall see in Chapter \(8,\) weak acids such as acetic acid only partially dissociate in solution, as shown in the following simplified net ionic equilibrium reaction. $$\mathrm{CH}_{3} \mathrm{COOH} \rightleftharpoons \mathrm{H}^{+}+\mathrm{CH}_{3} \mathrm{COO}^{-}$$ (a) Suppose that initially, only \(0.10 \mathrm{M}\) acetic acid is present. Analysis of the equilibrium mixture shows that the concentration of acetic acid reduces to 0.098 M. Determine the equilibrium concentrations of \(\mathrm{H}^{+}\) and \(\mathrm{CH}_{3} \mathrm{COO}^{-}\) (b) What is the expected equilibrium constant, \(K\), for this reaction?

Consider the reaction shown below at a certain temperature. $$2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)$$ The equilibrium constant, \(K\), is equal to \(8.7 \times 10^{3}\) at a certain temperature. At equilibrium, it is found that \\[ \left[\mathrm{H}_{2}\right]=1.9 \times 10^{-2} \mathrm{M} \text { and }\left[\mathrm{O}_{2}\right]=8.0 \times 10^{-2} \mathrm{M} . \text { What is } \\] the concentration of \(\mathrm{H}_{2} \mathrm{O}\) at equilibrium?

Write the equilibrium expression for this reaction: $$2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g)$$

Pure carbon exists is several forms, two of which are diamond and graphite. The conversion of the diamond form to the graphite form is exothermic to a very slight extent. How is it that jewelers can advertise "Diamonds are forever"?

The following reaction is endothermic. $$3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{O}_{3}(g)$$ If the reaction is at equilibrium, tell whether the equilibrium will shift to the right or the left if we: (a) Remove some \(\mathrm{O}_{3}\) (b) Remove some \(\mathrm{O}_{2}\) (c) Add some \(\mathrm{O}_{3}\) (d) Decrease the temperature (e) Add a catalyst (f) Increase the pressure

See all solutions

Recommended explanations on Chemistry Textbooks

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.