Answered

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Study the reactions.

C2H2 (g)+2H2 (g)→C2H6 (g)ΔH=−75 kcal
C2H4 (g)+H2 (g)→C2H6 (g)ΔH=−33 kcal
C2H2 (g)+H2 (g)→C2H4 (g)ΔH= ?

How can the enthalpy of Reaction III be calculated correctly?

A) by first multiplying the enthalpy of Reaction 2 by two and then finding the sum of the enthalpy of Reaction 1 and the enthalpy of Reaction 2

B) by first reversing Reaction 2 and then adding the enthalpy of Reaction 1 and the enthalpy of Reaction 2
C) by adding the enthalpy of Reaction 1 and the enthalpy of Reaction 2
D) by subtracting the enthalpy of Reaction 1 from the enthalpy of Reaction 2

Sagot :

Answer:

B) by first reversing Reaction 2 and then adding the enthalpy of Reaction 1 and the enthalpy of Reaction 2

Explanation:

C2H2 (g)+2H2 (g)→C2H6 (g)   ΔH=−75 kcal

C2H4 (g)+H2 (g)→C2H6 (g)     ΔH=−33 kcal

C2H2 (g)+H2 (g)→C2H4 (g)     ΔH= ?

First step - Reverse reaction 2.

C2H6 (g)  → C2H4 (g) + H2 (g)    ΔH= 33 kcal

Second step - Add reaction 1 and the reversed reaction 2. This gives us:

C2H2 (g) + 2H2 (g)  +  C2H6 (g) → C2H6 (g) + C2H4 (g) + H2 (g)

Simplifying gives:

C2H2 (g)+H2 (g)→C2H4 (g)

The correct option is:

B) by first reversing Reaction 2 and then adding the enthalpy of Reaction 1 and the enthalpy of Reaction 2

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