Question

How would the change in internal energy, ΔE, differ among these various reactions of methane gas with oxygen gas to produce carbon dioxide and steam?

Answer 1

Methane gas reacts with oxygen gas to produce gaseous carbon dioxide and water vapor. This reaction could occur in several different ways. The methane and oxygen could simply react directly to produce carbon dioxide and water vapor as shown in equation A.

A. 2 CH4 (g) + 4 O2 (g) → 2 CO2 (g) + 4 H2O (g)

Or, the methane gas could first be broken into its elements and the carbon dioxide and steam could be formed by reaction with O2 (equation b).

B. Step 1: 2 CH4 (g) → 2 C (s) + 4 H2 (g)
Step 2: 2 C (s) + 4 H2 (g) + 4 O2 (g) → 2 CO2 (g) + 4 H2O (g)
Overall reaction: 2 CH4 (g) + 4 O2 (g) → 2 CO2 (g) + 4 H2O (g)

Alternatively, two methane molecules could react to form ethane (C2H4) and hydrogen gas, which could then react with oxygen gas to form carbon dioxide and water vapor, as shown in equation C.

C. Step 1: 2 CH4 (g) → C2H4 (g) + 2 H2 (g)
Step 2: C2H4 (g) + 2 H2 (g) + 4 O2 (g) → 2 CO2 (g) + 4 H2O (g)
Overall reaction: 2 CH4 (g) + 4 O2 (g) → 2 CO2 (g) + 4 H2O (g)

Answer 2

OMG !!

Answer 3

what?

Answer 4

this is such a huge question i mean very long

Answer 5

ya. . . . .

Answer 6

i have to go through

Answer 7

It wouldn't differ, since internal energy is a state function just like enthalpy and entropy. You can get there any way you like, and the change in internal energy will be EXACTLY the same. This is the answer, and it really is that easy.

Another state function is temperature. If it was 50 degrees and now it's 100 degrees, it doesn't matter if you dropped down to 20 degrees before going up to 200 degrees and then went back down to 100 degrees to stop or just went directly from 50 to 100 degrees. The ΔT would be exactly the same in both cases, ΔT=50.