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.
