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CH4 (g) yields C(g) + 4H (g) (reaction for expansion)

Compare the reaction for the "expansion" of methane with the reverse of the reaction that represents the standard enthalpy of formation. Which properties are the same for both reactions and which are different?
1. the number of atoms of each element, 2. the state of the methane molecules, 3. the enthalpy change of the reaction, 4. the state of the carbon atoms.

User RoryWoods
by
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2 Answers

2 votes

Answer:

Same:

-the number of atoms

of each element

-the state of the

methane molecules

Different:

-the state of the

carbon atoms

-the enthalpy change

of the reaction

Step-by-step explanation:

The enthalpy changes are different for the two reactions even though all the same elements are involved in equal numbers. This illustrates that the enthalpy change of a reaction is dependent on the states of the reactants and products, as well as how the atoms are bonded.

User Amit Prakash
by
7.6k points
5 votes

Answer:

Here's what I find.

Step-by-step explanation:


\rm CH$_(4)$(g) $\, \rightleftharpoons \,$ C(g) + 4H(g)


\text{Same} =\begin{cases}1. & \text{The number of atoms of each element}\\2. & \text{The state of the methane molecules}\\4. &\text{The state of the carbon atoms}\\\end{cases}


\text{Different} =\begin{cases}3. & \text{The enthalpy change of the reaction}\\\end{cases}

The sign of ΔH changes when you reverse the reaction.

User Zlr
by
7.5k points