Question

Consider the following numbered processes: 1. A → 2B 2. B → C + D 3. E → 2D ΔH for the process A → 2C + E is

Answer 1

Answer:

ΔH = ΔH₁ + ΔH₂ - ΔH₃

Explanation:

Given that:

1. A → 2B

2. B → C + D

3. E → 2D

Assuming from the corresponding ΔH for process 1, 2 and 3 are ΔH₁, ΔH₂, ΔH₃ respectively.

To estimate the ΔH for the process A → 2C + E

We multiply 2 with equation 2 where (B → C + D)

2B → 2C + 2D ⇒ 2ΔH₂

Also, let's switch equation (3), such that we have,

2D → E -ΔH₃

The summation of all the equation result into :

A → 2C + E

where; ΔH = ΔH₁ + ΔH₂ - ΔH₃

Answer 2

Based on Hess law of constant heat summation, the ΔH for the process A → 2C + E is ΔH1 + ΔH2 - ΔH3

What is the enthalpy of the series steps of reaction steps?

According to Hess' law of constant heat summation, the heat of any reaction ΔH∘f for a specific reaction is equal to the sum of the heats of reaction for any set of reactions which in sum are equivalent to the overall reaction.

Thus, ΔH for the process A → 2C + E is calculated as follows:

• Reaction step 1: A → 2B ; ΔH1
• Reaction step 2: B → C + D ; ΔH2
• Reactipn step 3: E → 2D ; ΔH3

The A → 2C + E can be obtained from the summation of the processes above:

A ---> 2B

2B ---> 2C + 2D

2D ---> E by reversing reaction step 3

Thus, ΔH = ΔH1 + ΔH2 - ΔH3

Therefore, the ΔH for the process A → 2C + E is ΔH1 + ΔH2 - ΔH3

Learn more about enthalpy and Hess law at: https://brainly.com/question/11628413