C2H2(g) + 2H2(g) --> C2H6(g)

Information about the substances involved in the reaction represented above is summarized in the following table:

C2H2(g) 226.7 kj/mol
C2H6(g) -84.7 kj/mol
[the values are the change in enthalpy]

a. write the equation for the heat formation of C2H6.
b. Use the above information to determine the enthalpy of reaction for the equation given.

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A, for some reason, is throwing me off. I don't know what it's asking. Help?

Thanks!

Do part b first.

delta Hrxn = (delta H products) - (delta H reactants)

Then write the equation and add heat to the left side or to the right side depending upon the sign of delta Hrxn. I think that will be negative; therefore,
C2H2(g) + 2H2(g) --> C2H6(g) + heat

For part A, it is asking you to write the equation for the heat formation of C2H6. The heat formation (also called the enthalpy of formation) is the change in enthalpy when one mole of a substance is formed from its elements in their standard states. In this case, you need to write the balanced equation that represents the formation of C2H6 from its elements in their standard states.

The equation for the heat formation of C2H6 can be written as follows:

C2H2(g) + 3H2(g) --> C2H6(g)

In this equation, C2H2(g) and H2(g) are the elements in their standard states, and C2H6(g) is the substance being formed.

For part B, you need to calculate the enthalpy of reaction for the given equation. The enthalpy of reaction (also called the heat of reaction or ΔH) is the change in enthalpy for a chemical reaction. It represents the difference in energy between the reactants and the products.

To determine the enthalpy of reaction, you need to use the enthalpy values given for C2H2(g) and C2H6(g).

ΔH = Σ(ΔHf(products)) - Σ(ΔHf(reactants))

ΔH = [-84.7 kj/mol] - [226.7 kj/mol + 2(0)]

ΔH = -84.7 kj/mol - 226.7 kj/mol

ΔH = -311.4 kj/mol

The enthalpy of reaction for the given equation is -311.4 kj/mol. This means that the reaction is exothermic, as indicated by the negative sign.

For part a, you are being asked to write the equation for the heat formation (also known as heat of formation) of C2H6.

The heat of formation of a compound is the change in enthalpy when one mole of the compound is formed from its constituent elements in their standard states.

In this case, C2H6 is formed from its constituent elements C2H2 and H2. Therefore, the equation for the heat formation of C2H6 can be written as:

(1/2)C2H2(g) + H2(g) → C2H6(g)

Note that the coefficient of C2H2 is (1/2) because the standard enthalpy change given for C2H2 is in kilojoules per mole, and the equation is balanced in terms of moles.

Moving on to part b, you are asked to determine the enthalpy of reaction for the given equation:

C2H2(g) + 2H2(g) → C2H6(g)

To find the enthalpy of reaction, you need to consider the difference in enthalpy between the products and the reactants.

The given enthalpy values are in kilojoules per mole and represent the change in enthalpy for each substance involved.

To determine the enthalpy of reaction, you need to calculate the sum of the enthalpy of the products minus the sum of the enthalpy of the reactants:

Enthalpy of reaction = [Enthalpy of C2H6(g)] - [Enthalpy of C2H2(g) + Enthalpy of H2(g) + Enthalpy of H2(g)]

Enthalpy of reaction = [-84.7 kj/mol] - [226.7 kj/mol + 2(0 kj/mol)]

Enthalpy of reaction = -84.7 kj/mol - 226.7 kj/mol

Enthalpy of reaction = -311.4 kj/mol

Therefore, the enthalpy of reaction for the given equation is -311.4 kilojoules per mole.