Answer:
B. CH4(g) + H₂O(g) → CO(g) + 3H₂(g)
Step-by-step explanation:
The increase in the number of moles of gas is a good indicator of an increase in entropy. Therefore, the reaction that produces more moles of gas than it consumes will have an increase in entropy.
A. 2H₂S(g) + 30₂(g) → 2H₂O(g) + 2SO2(g)
In this reaction, the number of moles of gas on the left side is 3, and on the right side, it is also 3. Thus, there is no change in the number of moles of gas, and there is no increase in entropy.
B. CH4(g) + H₂O(g) → CO(g) + 3H₂(g)
In this reaction, the number of moles of gas on the left side is 2, and on the right side, it is 4. Thus, there is an increase in the number of moles of gas, and there is an increase in entropy.
C. 2NO(g) + 2H₂(g) → N₂(g) + 2H₂O(g)
In this reaction, the number of moles of gas on the left side is 4, and on the right side, it is 3. Thus, there is a decrease in the number of moles of gas, and there is no increase in entropy.
D. 2CO(g) + 2NO(g) → 2C0₂(g) +N₂(g)
In this reaction, the number of moles of gas on the left side is 4, and on the right side, it is also 4. Thus, there is no change in the number of moles of gas, and there is no increase in entropy.
Therefore, the only reaction that has an increase in entropy is B. CH4(g) + H₂O(g) → CO(g) + 3H₂(g).
Hope this helps!