Pls help me with this chemistry work.

Pls Help Me With This Chemistry Work.

Answers

Answer 1
Answer:  −2430 kJ

Explanation:

This seems to be a Hess' Law problem. Hess' Law states that "states that regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes." (Libretext)

Compare the step reaction to the final reaction to see the similarity and differences. We may have to flip an equation or multiply it by a constant to obtain the number of moles of the species in the final equation. Just remember, whatever we do to the equation, we have to do to the ΔH as well.

3C  +  4H₂  →  C₃H₈      ΔH= +104 kJ  [1]

      C + O₂ → CO₂        ΔH= -394 kJ  [2]

  2H₂ + O₂ → 2H₂O      ΔH= -572 kJ  [3]

C₃H₈  + 5O₂  → 3CO₂ + 4H₂O  ΔH = ?

_____________________________________________

Compare the Equation

Equation [1]

In [1] the C₃H₈is on the product side. But the final equation has the C₃H₈ on the reactant side. Thus we have to flip the equation. This flips the sign of ΔH.

                C₃H₈   →   3C  +  4H₂     ΔH= - 104 kJ  [1]

Equation [2]

The CO₂ in the final equation is on the product side but is 3 moles. Thus we have to multiply [2] by three. The ΔH is also multiplied by 3.

               3C + 3O₂ → 3CO₂        ΔH= 3 (-394 kJ)    [2]

Equation [3]

The H₂O in the final equation has a coefficient of 4. Thus we have to multiply [3] by 2.

              4H₂ + 2O₂ → 4H₂O      ΔH= 2(-572 kJ)  [3]

Combine the Equations

Combine the newly sorted versions of the equation and simplify to see if it matches the final equation. We also add the ΔH.

⇒  C₃H₈ +  3C + 3O₂ + 4H₂ + 2O₂   →   3C  +  4H₂  + 3CO₂ + 4H₂O

              ΔH= - 104 kJ + 3 (-394 kJ) + 2(-572 kJ)

⇒   C₃H₈ +  + 5 O₂   →  3CO₂ + 4H₂O

              ΔH= - 104 kJ + 3 (-394 kJ) + 2(-572 kJ) = −2430 kJ

∴  the  ΔH of the final equation is  −2430 kJ.

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HOPE IT'S HELP

FOLLØW ME PLEASE THANKS IN ADVANCE

Heart my answer if it's helpful

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