The third law of thermodynamics states: A) The entropy of the universe is increa

Question

The third law of thermodynamics states: A) The entropy of the universe is increasing. B) The entropy of the universe is constant. C) The entropy is zero at 0 K for a perfect crystal. increasing temperature, D) The absolute entropy of a substance decreases with of system and E) The entropy of the universe the sum of the entropy surroundings For which of the following processes would delta S be expected to be most positive? O_2(g) + 2H_2(g) rightarrow 2H_2O(g) H_2O(l) rightarrow H_2O(s) NH_3(g) + HCl(g) rightarrow NH_4Cl(g) 2NH_4NO_3(g) rightarrow 2N_2(g) + O_2(g) + 4H_2O(g) N_2O_4(g) rightarrow 2NO_2(g) Consider the reaction Calculate delta S for the reaction. A) 809.08 JK B) 89.38 JK C) 453.76 J/K D) -265.94 JK. E) 1164.40 U/K Calculate AH for the reaction. A) 110.022 B) 10.572 C) 121.311 kJ D) 21.861 kJ E) 155.178 kJ

Explanation / Answer

15)

Answer: C

16)

So is maximum positive when product has maximum number of gases and reactant has least or none

In option D we can see than product has 7 moles of gas but reactant has 0

So,

So is maximum positive for D

Answer: D

17)

Balanced chemical equation is:

2N2O5(g) ---> 4NO2(g) + 1O2(g)

So rxn = 4*Sof(NO2(g)) + 1*Sof(O2(g)) - 2*Sof( N2O5(g))

So rxn = 4*(239.9) + 1*(204.8) - 2*(355.32)

So rxn = 453.76 J/k

Answer: C

18)

Balanced chemical equation is:

2N2O5(g) ---> 4NO2(g) + 1O2(g)

Ho rxn = 4*Hof(NO2(g)) + 1*Hof(O2(g)) - 2*Hof( N2O5(g))

Ho rxn = 4*(31.15) + 1*(0.0) - 2*(11.289)

Ho rxn = 102.02 KJ

None of the option matches

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