/*! 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 20 Write the chemical equation for ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Write the chemical equation for the formation reaction of \(\mathrm{H}_{2} \mathrm{~S}(g)\).

Short Answer

Expert verified
The formation reaction is \( \text{H}_2(g) + \frac{1}{8}\text{S}_8(s) \rightarrow \text{H}_2\text{S}(g) \).

Step by step solution

01

Understand the Chemical Components

In this formation reaction, we need to produce hydrogen sulfide gas, which is represented by the chemical formula \( \text{H}_2\text{S}(g) \). This involves hydrogen (\( \text{H} \)) and sulfur (\( \text{S} \)) as the reactants.
02

Identify the States of Each Element

In a formation reaction, elements must be in their standard states. Here, hydrogen is a diatomic molecule \( \text{H}_2(g) \) in its gaseous state and sulfur is usually found as solid sulfur, \( \text{S}_8(s) \).
03

Write the Formation Reaction

For the formation of \( \text{H}_2\text{S}(g) \), hydrogen gas and sulfur solid react to form hydrogen sulfide gas. The chemical equation can be written as: \[ \text{H}_2(g) + \frac{1}{8}\text{S}_8(s) \rightarrow \text{H}_2\text{S}(g) \]
04

Balance the Chemical Equation

The equation is already balanced in terms of hydrogen. For sulfur, adjust it to produce the desired compound. Hence, the complete balanced equation is: \[ \text{H}_2(g) + \frac{1}{8}\text{S}_8(s) \rightarrow \text{H}_2\text{S}(g) \]

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.

Formation Reaction
A formation reaction is a type of chemical reaction where one mole of a compound is formed from its constituent elements in their standard states. This means that the starting materials, or reactants, are the pure elements that make up the compound. Understanding this concept is key to writing correct chemical equations for these types of reactions. In the case of hydrogen sulfide (\(\text{H}_2\text{S}(g)\)), we begin with hydrogen and sulfur. Because these elements must be in their standard states, hydrogen is present as diatomic hydrogen gas \(\text{H}_2(g)\), and sulfur is in its elemental form \(\text{S}_8(s)\), which is the most stable form of sulfur at room temperature.
Hydrogen Sulfide
Hydrogen sulfide is a chemical compound with the formula \(\text{H}_2\text{S}\). It is a colorless gas well-known for its distinctive smell of rotten eggs. This simple binary chemical compound is composed of two hydrogen atoms bonded to a single sulfur atom. While it occurs naturally in crude petroleum, natural gas, and hot springs, it is also produced by the bacterial breakdown of organic matter in the absence of oxygen.

Here are some key points about hydrogen sulfide:
  • It is denser than air, which means it can accumulate in low-lying areas.
  • It is flammable and can cause irritation to the eyes, nose, and throat.
  • Despite its toxicity at higher concentrations, humans can detect it at very low concentrations due to its strong odor.
Balancing Chemical Equations
Balancing chemical equations involves ensuring equal numbers of each type of atom on both sides of the equation. This reflects the conservation of mass, meaning that matter cannot be created or destroyed in a chemical reaction. When writing a balanced chemical equation for the formation of hydrogen sulfide, we start with the unbalanced equation:\[\text{H}_2(g) + \text{S}_8(s) \rightarrow \text{H}_2\text{S}(g)\]

To balance this, we need to make the number of each type of atom the same on both sides. Since sulfur exists naturally as \(\text{S}_8\), we adapt the equation by using \(\frac{1}{8}\text{S}_8\), balancing the sulfur atoms. Thus, the fully balanced equation is:\[\text{H}_2(g) + \frac{1}{8}\text{S}_8(s) \rightarrow \text{H}_2\text{S}(g)\]

Balancing chemical equations is essential not only for correctly representing chemical reactions but also for calculating the amounts of reactants and products involved in reactions.

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

Is the following reaction the appropriate one to use in determining the enthalpy of formation of methane, \(\mathrm{CH}_{4}(g) ?\) Why or why not? $$ \mathrm{C}(g)+4 \mathrm{H}(g) \longrightarrow \mathrm{CH}_{4}(g) $$

How fast (in meters per second) must an iron ball with a mass of \(56.6 \mathrm{~g}\) be traveling in order to have a kinetic energy of \(15.75 \mathrm{~J} ?\) The density of iron is \(7.87 \mathrm{~g} / \mathrm{cm}^{3} .\)

Hydrogen sulfide, \(\mathrm{H}_{2} \mathrm{~S}\), is a poisonous gas with the odor of rotten eggs. The reaction for the formation of \(\mathrm{H}_{2} \mathrm{~S}\) from the elements is $$ \mathrm{H}_{2}(g)+\frac{1}{8} \mathrm{~S}_{8}(\text { rhombic }) \longrightarrow \mathrm{H}_{2} \mathrm{~S}(g) $$ Use Hess's law to obtain the enthalpy change for this reaction from the following enthalpy changes: $$ \begin{gathered} \mathrm{H}_{2} \mathrm{~S}(g)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{SO}_{2}(g) ; \Delta H=-518 \mathrm{~kJ} \\\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g) ; \Delta H=-242 \mathrm{~kJ} \\ \frac{1}{8} \mathrm{~S}_{8}(\text { rhombic })+\mathrm{O}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g) ; \Delta H=-297 \mathrm{~kJ} \end{gathered} $$

A bullet weighing 245 grains is moving at a speed of \(2.52 \times 10^{3} \mathrm{ft} / \mathrm{s} .\) Calculate the kinetic energy of the bullet in joules and in calories. One grain equals \(0.0648 \mathrm{~g}\).

Phosphoric acid, \(\mathrm{H}_{3} \mathrm{PO}_{4}\), can be prepared by the reactio of phosphorus(V) oxide, \(\mathrm{P}_{4} \mathrm{O}_{10}\), with water. $$ \frac{1}{4} \mathrm{P}_{4} \mathrm{O}_{10}(s)+\frac{3}{2} \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{3} \mathrm{PO}_{4}(a q) ; \Delta H=-96.2 \mathrm{~kJ} $$ What is \(\Delta H\) for the reaction involving \(1 \mathrm{~mol}\) of \(\mathrm{P}_{4} \mathrm{O}_{10}\) ? $$ \mathrm{P}_{4} \mathrm{O}_{10}(s)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 4 \mathrm{H}_{3} \mathrm{PO}_{4}(a q) $$

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.