1. 1 50137,5 . ∆ 4.
.
1. , , (, 4 16 /):
4 n :
n 4= 
=62,5 .
2. , 1 :
3. :
CH4() + 22() = C2() + 22(), Q =802,2 ,
Q =− ∆ =(∆ 2() +2 ∆ 2())−(∆ 4() +2 ∆ 2())=
=[(−393,5)+2(−241,8)−(∆ 4())−2(0)]=−802,2 .
: ∆ 4=(802,2−393,5−483,6)=−74,9 / 4.
2. , 100 , .., 2() 2().
.
1.
26() + ,5O2() = 2CO2() + 32(); ∆ =−1559,87 .
2. , (..), (, 1 .. , 22,4 ):
n :
n 26()= 
=4,46 .
3. (∆ 298) , (. . 2) 26():
∆ =(2 ∆ O2() +3 ∆ 2O ())−(∆ 26() +3,5 ∆ O2())=
= [2(−393,5)+3(−241,8)]−[−84,7+3,5×0]=−1427,7 / 26().
Q =− ∆ =1427,7 .
4. , .. 4,46 (100 , ..):
Q =− ∆ .=1427,7×4,46=5767,42 .
3. , . .
1) 2N3() = N2() + 3H2(),
2) NH4NO3() = N2O() + 22(),
3) 22() + O2() = 22O(),
4) 22() + O2() = 22O().
. . , , . , .
2- , 4 , , , S 1>0.
(∆S 298) :
∆S 298= S N2()+3 S 2()−2 S N3()=191,5+ 3×130,5−2×192,7=197,6 /.
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1 3 , , ∆S 2>0. (∆S 298) :
∆S 298= S N2()+2 S 2() S N4NO3() =219,8+2×188,7−151=446,2 /.
(3) (4) , , ∆S 3<0 ∆S 4<0, D S 4 , .. ∆S 3, S 2() > S 2().
4. 2500 ( ∆H ∆S ):
TiO2() + 2() = Ti() + 2().
. , , (∆G <0).
1. ∆ G (. . . 2):
∆ G =(2∆ G ()+∆ G Ti())−(∆ G TiO2() +2 ∆G ()) =
= [2(−137,1)+0]−[(−888,6)−2×0] =614,4 .
∆ G >0, 298 .
2. 2500 , :
∆G =∆ H p− T∆S p.
∆H p ∆S p :
∆H p=(2 ∆H () + ∆H Ti())−(∆H TiO2()+2 ∆H ()) =
= [2(−110,5)+0]−[(−943,9)−2×0 ]=722,9 =722900 .
∆S p=(2 S ()+ S Ti())−(S TiO2()+ S ())=
= [2×197,5+30,6]−[50,3-2×5,7 ]=363,9 /.
∆ G= ∆H − T∆ S=722900−2500×363,9=−186850 .
∆G 2500=−186,85 .
∆G 2500< 0, 2500 .
5. , :
2 NO2() 
N2O4(),
: ∆ =−55,3 ; ∆S =−175,8 /.
. , . , ∆G =0. :
∆G = ∆H − T∆ S=0.
:
=314,56 (41,4).
41,4 .
