Calculate the entropy change when 3.66 g of h2 reacts with o2 according to the reaction 2 h2(g) + o2(g) → 2 h2o(ℓ) at 298 k and 1 atm pressure. the standard molar enthalpy of formation of h2o(ℓ) at 298 k is −285.8 kj/mol. the corresponding free energy of formation is −237.2 kj/mol. answer in units of j/k.

The entropy change will be ΔS = - 150.0 J/K.

Explanation:

2 H₂(g) + O₂(g) → 2 H₂O(ℓ)  

ΔH = -258.8 kJ/mol

ΔG = - 237.2 kJ/mol

1 mol H₂ 2 g

x             3.66 g

x = 1.83 mol

1 mol H₂ 1 mol H₂O

1.83 mol H₂ 1.83 mol H₂O

ΔH = 1.83 mol x -258.8 kJ/mol

ΔH = - 478.78 kJ

ΔG = 1.83 mol x  - 237.2 kJ/mol

ΔG = 434.076 kJ

ΔG = ΔH - TΔS

ΔS = (ΔH - ΔG) / T

ΔS = [ - 478.78 kJ - (- 434.076 kJ)]/298 K

ΔS = - 150.0 J/K

they all get energy from eating other animals, dead animals to be more precise.

hienas tends to steal pray from other animals. crows and vultures tend to eat dead animals and leftovers.

hope it ,

bioteacher101

a). in this case, dh> 0 and sd> 0. if t becomes larger, the term "-tds" will become more negative. at some point, it is possible that -dts will be more negative than the dh is positive, and the reaction will become spontaneous. this is probably the correct answer, but you should look at all of the other possibilities and make sure of that.