contestada

Which of the following processes have a ΔS > 0? A. N2(g) + 3 H2(g) → 2 NH3(g) B. Na2CO3(s) + H2O(g) + CO2(g) → 2 NaHCO3(s) C. CH3OH(l) → CH3OH(s) D. All of these processes have a ΔS > 0. E. CH4(g) + H2O (g) → CO(g) + 3 H2(g)

Respuesta :

baraltoa

Answer:

ΔS > 0 only for choice E: CH4(g) + H2O (g) → CO(g) + 3 H2(g)

Explanation:

Our strategy in this question is to use the trend in entropies :

S (solids)  less than S (liquids) less than S (gases)

Also we have to look for the  molar quanties involved of each state and their change to answer the question:

A. N2(g) + 3 H2(g) → 2 NH3(g)

Here we have 4 moles gases going to 2 moles of products, so the change in entropy is negative.

B. Na2CO3(s) + H2O(g) + CO2(g) → 2 NaHCO3(s)

The change in entropy is negative since we have 2 mol gases in the reactants and zero in the products.

C. CH3OH(l) → CH3OH(s)

A liquid has a higher entropy than a solid so ΔS is negative

D. False see A,B,C

E. The change in moles of gases is 4 - 2= 2, therefore  ΔS is greater than O.

pstnonsonjoku

The reaction CH4(g) + H2O (g) → CO(g) + 3 H2(g) will have ΔS > 0.

The term entropy refers to the degree of disorder of a system. Hence, the change in entropy is positive (greater than zero) when there is an increase in the degree of disorderliness of the system.

As such, the reaction;

  • CH4(g) + H2O (g) → CO(g) + 3 H2(g)

will experience an increase in entropy since there is an increase in the number of molecules of  gaseous species from left to right.

Learn more about entropy: https://brainly.com/question/1217654

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