Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Methane, \(\mathrm{CH}_{4}(g),\) burns in oxygen, \(\mathrm{O}_{2}(g),\) to produce carbon dioxide, \(\mathrm{CO}_{2}(g),\) and water, \(\mathrm{H}_{2} \mathrm{O}(l) .\) a. Write a balanced equation for this process. b. This process is described as irreversible. What does this mean?

Short Answer

Expert verified
a. The balanced equation is: \(\mathrm{CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)}\). b. This process is irreversible because the products do not readily reform the reactants.

Step by step solution

01

Write the unbalanced chemical equation

Start by identifying the reactants and products. Methane (\(\mathrm{CH_4(g)}\) and oxygen (\(\mathrm{O_2(g)}\) are the reactants, and carbon dioxide (\(\mathrm{CO_2(g)}\) and water (\(\mathrm{H_2O(l)}\) are the products. The unbalanced equation is:\[\mathrm{CH_4(g) + O_2(g) \rightarrow CO_2(g) + H_2O(l)}\]
02

Balance carbon and hydrogen atoms

Look at the number of each type of atom on both sides. There is 1 carbon atom in methane and 1 in carbon dioxide; thus carbon is balanced. Next, there are 4 hydrogen atoms in methane and 2 in one water molecule, so balance hydrogen by multiplying water by 2:\[\mathrm{CH_4(g) + O_2(g) \rightarrow CO_2(g) + 2H_2O(l)}\]
03

Balance oxygen atoms

Now count the oxygen atoms. On the product side, there are 2 from carbon dioxide and 2×1=2 from the water, totaling 4 oxygen atoms. Thus, balance the oxygen by placing a coefficient of 2 before the molecular oxygen:\[\mathrm{CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)}\]
04

Verify the balanced equation

Double-check the atom counts: 1 carbon, 4 hydrogen, and 4 oxygen atoms on both sides confirm the equation is balanced.
05

Explain irreversibility

An irreversible process in chemistry generally means the reaction, once complete, cannot easily return to the original reactants under normal conditions. For a combustion reaction like this, the products are stable and cannot convert back to methane and oxygen spontaneously.

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 Vaia!

Key Concepts

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

Combustion Reaction
Combustion reactions are a type of chemical reaction where a substance reacts with oxygen to produce heat and light. This process usually involves hydrocarbons, which are compounds made of hydrogen and carbon, just like methane.
In a combustion reaction, the main products are often carbon dioxide and water. The type of combustion discussed in the original exercise is a complete combustion since all the carbon and hydrogen in methane react fully with oxygen.
Complete combustion happens when there is enough oxygen present. It results in a clean reaction with no soot or carbon monoxide, which are side products of incomplete combustion.
If you’ve ever lit a match, driven a car, or seen a candle burn, you have witnessed a combustion reaction!
  • Reacts with oxygen
  • Produces heat and light
  • Results in carbon dioxide and water
Balancing Chemical Equations
Balancing chemical equations is essential for reflecting the conservation of matter. In any chemical reaction, matter cannot be created or destroyed, which means that the mass of the reactants must be equal to the mass of the products.
To balance an equation, make sure that the number of each type of atom on the left side of the equation (reactants) equals the number on the right side (products).
Steps to balance chemical equations include:
  • Identify reactants and products
  • Count atoms of each element in reactants and products
  • Use coefficients to balance each element on both sides
In the methane combustion reaction, the equation was balanced by changing the coefficients, not the subscripts. This simple approach keeps the molecules intact, which is crucial for accurate representation of chemical processes.
Irreversibility in Chemistry
An irreversible process in chemistry is one that proceeds in one direction only and cannot be easily undone. In the context of chemical reactions, irreversibility means that the products will not spontaneously turn back into the reactants under normal conditions.
In the original exercise, the burning of methane is irreversible because the products, carbon dioxide and water, are very stable and do not react to reform methane and oxygen easily.
Irreversibility is common in combustion reactions due to:
  • The release of energy
  • Stable product formation
  • High temperature conditions
These factors make it difficult for the reverse process to occur, ensuring that the reaction's progress is in one direction only. This concept is crucial for understanding why some reactions, once completed, are permanent.
Oxygen as a Reactant
Oxygen is a vital reactant in combustion and many other chemical reactions. It is highly reactive and readily combines with other elements, particularly carbon and hydrogen as seen in the combustion of methane.
Oxygen plays the role of an oxidizing agent. It accepts electrons during the reaction, which is part of the oxidation process. This characteristic of oxygen is what drives the reaction forward energetically, resulting in the release of heat and light.
Key points to remember about oxygen in reactions are:
  • It acts as an oxidizing agent
  • Combines with carbon and hydrogen
  • Drives energy release
Without oxygen, reactions like burning wood or fuel in engines would not be possible. It's essential for various industrial, everyday processes and even biological metabolism.

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

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.

Sign-up for free