Calculate the entropy change for the reaction: n2(g) + 3h2(g) -> 2nh3(g) entropy data: nh3 = 192.5 j/mol k h2 = 130.6 j/mol k n2 = 191.5 j/mol k

ΔS⁰ = -198.3 J/K

Explanation:

N₂(g) + 3H₂(g)  → 2NH₃(g)

S⁰:   N₂(g) = 1mole(191.5J/mole·K) = 191.5J/K

S⁰: 3H₂(g) = 3moles(130.6J/mole·K) = 391.8J/K                                

S⁰: 2NH₃(g) = 2moles(192.5J/mole·K) = 385J/K

ΔS⁰ = ∑n·S⁰(Products) - ∑n·S⁰(Reactants

      =[385J/K] - [191.5J/K + 391.8J/K]

      = (385 - 191.5 - 391.8)J/K

      = -198.3J/K

h let me think

1. a)c-14 has two more protons

2. d)they have different numbers of protons, neutrons, and valence electrons; they are all gases at room temperature.

3. c) iron and nickel.