please help and explain... I am beyond lost on this assignment 1. A certain proc
ID: 893018 • Letter: P
Question
please help and explain... I am beyond lost on this assignment
1. A certain process has H>0 and S<0. Which of the following is a correct conclusion about this process
Non spontaneous at all T.
Spontaneous at high T / Spontaneous at low T / Spontaneous at all T / None of the above
2. the formation constant for the reaction Ag+(aq)+2NH3(aq) ß-à Ag(NH3)2+(aq) is K=1.7x10^7 at 25 deg C. what is G (kJ/mol) at this temp
-23 / -18 / -3.5 / -1.5 / -41
3. Nitric oxide reacts with chlorine to form NOCL. What is the free energy (kJ) of the reaction at 277 dec C.
2NO(g) + Cl2(g) à 2NOCl(g)
+144 / -41.0 / -10.3 / -22.2 / +41.0
4. How much free energy is released or absorbed(kJ) when 95.0 g of ozone is produced.
3 O2(g)à2 O3(g) G = +326 kJ
+645 / -323 / +95 / +323 / -645
5. Estimate the boiling point (deg C) of phosphoric acid? H3PO4(s)ßàH3PO4(l)
42 / 181 / 315 / -92 / 305
6. Calculate the S (J/K) for the reaction, SiCl4(g)+2Mg(s)à2MgCl2(s)+Si(s)
+198 / -198 / +255 / -255 / +472
7. Calculate the equilibrium constant, K, for the following reaction at 25 deg C: CH4(g)+2H2O(g)ßà CO2(g) G =-113.6kJ
Which of the following results in a decrease in the entropy of the system
O2(g),300KàO2(g), 400K / H2O(s)àH3O(l) / N2(g)àN2(aq) / NH3(l)àNH3(g) / 2H2O2(g)à2H2O(g)+O2(g)
8. Identify the incorrect statemtent?
When G >0, energy can be added to force a non-spontaneous to proceed.
When K>1, G <0.
Enthalpy is a good predictor of free energy.
Entropy of the universe increases in a spontaneous reaction
All of the above
Explanation / Answer
1. DH = + ve , Ds = -Ve
DG = DH-TDS
So that net DG = +ve.( Non spontaneous Non spontaneous at all T.)
2. Ag+(aq)+2NH3(aq) --> Ag(NH3)2+(aq) K=1.7x10^7 at 25 deg C.
DG = -2.303RTlogK
= -2.303*8.314*298log(1.7*10^7)
= -41.25 kj/mol
3. 2NO(g) + Cl2(g) ---> 2NOCl(g)
DGrxn = G products - G rectants.
= (2* 66.08) - (2*87.6 + 0)
= -43.04 kj/mol
4. 3 O2(g)---> 2O3(g) Dg = +326 KJ
nO of moles of O3 = 95/48 = 1.98 mole
from equation 2 mole O3 = 326 kj
so that
1.98 mole O3 = 326*1.98/2 = +322.74 kj
answer : +323 kj
5. no data
6. SiCl4(g)+2Mg(s) ----> 2MgCl2(s)+Si(s)
DS rxn = S products - S reactant
= (2*89.6+18.8) -(2*32.7+330.9)
= -198.3 J/K mol
answer : -198.3 J/K mol
7.
DG = -2.303RT log K
-113*1000 = -2.303*8.314*298 log K
k = 6.37*10^19
decrease in the entropy of the system : O2(g),300KàO2(g), 400K
8.
Enthalpy is a good predictor of free energy.
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