Scoring: your score will be based on the number of correct matches minus the mumber of incorrect matches. there is no penalty for missing matches use the references to secess important values if seeded far this question predict whether as for cach reaction would be greater than zero, less than zero, or too close to zero to decide. clear all i(g) cl2(g) 21ci(g) pci5(g pch(g)+ cl2(g) as 0 co2(g) h2(g)- +h,0(g) co(g) as< 0 2co(g)+ 2no(g) 2co; (g) n() too close to decide 2h2o2()2h2o(d) o2(g)

a. There is little or no change in the entropy.

b. ΔS is greater than zero.

c. There is little or no change in the entropy.

d. ΔS is less than zero.

e. ΔS is greater than zero.

Explanation:

The change in the entropy (ΔS) depends on the change in the amount of gaseous moles, Δn(g) = ngas, products - ngas,reactants.

a. I₂(g) + Cl₂(g) → 2 ICl(g)

Δn(g) = 2 - 2 = 0. There is little or no change in the entropy.

b. PCl₅(g) → PCl₃(g) + Cl₂(g)

Δn(g) = 2 - 1 = 1. ΔS is greater than zero.

c. CO₂(g) + H₂(g) → H₂O(g) + CO(g)

Δn(g) = 2 - 2 = 0. There is little or no change in the entropy.

d. 2 CO(g) + 2 NO(g) → 2 CO₂(g) + N₂(g)

Δn(g) = 3 - 4 = -1. ΔS is less than zero.

e. 2 H₂O₂(l) → 2 H₂O(l) + O₂(g)

Δn(g) = 1 - 0 = 1. ΔS is greater than zero.

the answer for sodium is 0.186

the answer for rubidium is 0.251