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Which reaction has a positive change in entropy? Reaction 1 Reaction 2 Reaction

ID: 1068727 • Letter: W

Question

Which reaction has a positive change in entropy? Reaction 1 Reaction 2 Reaction 3 All 3 Reactions What is the change in entropy of the surroundings, delta for Rea S_sarr degree, for reaction 1 at 25 degree C? -3.840 J/K -0.322 J/K -3.84 J/K -322 J/K Is Reaction 1 spontaneous at 25 degree C? No yes Insufficient information At what temperature will Reaction 3 become non-spontaneous? At temperatures below 462 degree C At temperatures below 0.46 K At temperatures above 189 degree C At temperatures above 735 K Reaction 3 is an equilibrium reaction: 3 H_2(g) + N_2(g) - 2 NH_3(g), Which of the following is a true statement concerning the spontaneous and the reverse reaction is spontaneous. The forward reaction is spontaneous and the reverse reaction in non-spontaneous. The forward and the reverse reactions are both spontaneous. The forward and the reverse reactions are both-spontaneous. In the industrial synthesis of ammonia (the haber process), reaction 3 is carried out at a temperature of 500 degree C, and at the following partial pressures: P_H2 = 2.5 atm, P_N2 atm, P_NH3 = 0.17 atm. What is the free energy change of the forward reaction under these non-standard conditions? -32 kj -47 kj -62 kj -85 kj

Explanation / Answer

7) entropy change

Sreaction = Sproducts Sreactants

reaction 1).   Sreaction = (239 J/mol.K) - (2x131 J/mol.K + 192 J/mol.K) = -215 J/mol.K

reaction 2) Sreaction = (2x193 J/mol.K) - (239 J/mol.K + 131 J/mol.K) = 16 J/mol.K

reaction 3) Sreaction = (2x193 J/mol.K) - (2x131 J/mol.K + 192 J/mol.K) = -68 J/mol.K

answer is 2 nd reaction change in entropy is positive

8) Ssurrounding = (q surrounding / Tsurrounding ) = (- qsystem /Tsurrounding)

qsystem is the same as enthalpy change and internal energy (internal energy change) of the system at constant pressure and constant volume.

Ssurrounding = (-96000 J / 298 K) = -322.14 J/K

option d is correct

9) for spontaneous  G < 0

G = H - T S > 0 for spontaneus

= 96000 J/mol - 298 K (-215 J/mol.K)

G = 160070 J/mol > 0 so reaction is non-spontaneous

and also S is negative so 1st reaction is non- spontaneous.

10) G = H - T S > 0 for non spontaneus

G = (-46000 J/mol) - T (-68 J/mol.K)

G = -46000 J/mol + T 68 J/mol.K

T > 676.47 K for non spontaneous

option D is correct.

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