MATHEMATICAL PROBLEM PAGE 11-15 OLUTIONS of the following problems will he fiven in the next issue. PROBLEMS
S
How many grams of mercuric oxide are required to produce sufficient oxygen to oxidize 10 g. of aluminum? The equilibrium constant, K, for the reaction Hz I2 2HI is given by the expression
+
K =
4xP
(a
- 2) ( b - x)
where a and b represent the initial concentrations of hydrogen and iodine and 22 represents the concentration of hydriodic acid formed a t equilibrium. What is the value of x in terms of a, 6, and K? A rotation of +0.235' was obtained when 0.200 g. of a certain optically active material was dissolved in 20 cc. of chloroform and placed in a tube 20 cm. long. What is the specific rotation? (The measurement was made a t 20°C. with yellow sodium light corresponding to the D line.) [ J . Am. Chem. SOG., 55,709 (1933).] The influence of temperature, T , on a chemical equilibrium constant, K,, is given by the relation
2.83; or, log w = 3/2 log 2 = 3/2 X 0.3010 = 0.4515. w = antilog 0.4515 = 2.83. 3. lo-'.' = x. x = antilog (-4.7) = antilog 5.3 = 2 X 1 -1 0 - = 0.000020; or, x = =%OX antilog 4.7 5 X lo-' 10-6. 1 1 4. (a) Wave-length = - 42,940 = 2.329 wave number 10-5 an. (b) 1A.U. = Em. 2.329 X lo-& an. = 2329 A.U. ( c ) I mp = mm. = lo-' cm. 2.329 X 10-6 em. = 232.9 mp. velocity - 3 X loT0cm. per sec. (d) frequency = wave-length 2.329 X cm. = 1.288 x per sec. ( e ) s (energy of quantum) = k (Plan&'s constant) x ~(frequency)= (6.547 X 10-27erg-second)X (1.288 X erg. 10'5 sec-1) = 8.434 X 10WLP 3.88 - 2.32 5. (a) Slope =- Y = xz - 2, (188.0 - 170.5) X 1.56 X 10%= 8g20, 17.5 (b) E = 2.303 x 1.987 X 8920 = 40.900 calories. (The value 41,500 was obtained from the original graph. The use of the small printed graph is less accurate.)
= se-"RT. 0LW22 = se 36.4 10 2303 = X 10-15.8,
6.
0.0 1.0
C 'I
0.02 1.00051
7
1-10 10, 249 (1933) 1
SOLUTIONS OF PROBLEMS
[J. CAEM.EDUC.,
X 2.54 = Volume of water = 100.0W2 X 2.54 = 2.54 X 10'O cc. 2'54 lo'' = 5.08 X 1011 drops. 0.05 2.54 x 10'O 2.20 = 27,940 tons, (b) 1000 X 2000 2'54 lo'' X 6.06 X 102' = 8.56 X 10" molecules. (') 18
(a)
x
41,500 - 1.987 x 563
= 1.4 X loLa X 10-16.1 = 1.4 X 8 X lo-' = 0.00112. 1.4 X Table I1 of the reference eves experimental values of (1.11, 1.08, and 1.11) x lo-=.
k
X lo'' X
=
=
8. k = s e - q n = . k = 1.4 X
I, k = I n s
loL8X
+(-&) ln
41,SOU --1.987T
e
.
e =
E .
ln s - RT -
In k = 30.275
500. - 41 L1.987T
This form of the logarithmic equation is preferred because it gives at a glance the energy of activation, E. For computational purposes it may be simplified to give 9069, log k = 13.146 - T g, v== KO. n log v = loge +log K. 1 1 log v = - log e + - log K (equation for straight line).
0.2 0.99968
determine the constants A and B in the equation = + + Bc by the method of difference quotients described by DeVries in J. CHEM.EDUC., 9, 2090 (1932).
X
s= O10-18.8 E O = 0.00220 X 6.3 X 10" = 1.39 X lOls. (The averaged value given in the reference is 1.4 X 10's.)
+ +
+
=
se-36.4
-
dlnKp dT RP
where A H is the heat of reaction and R is the gas constant. (a) Solve for in K, when A H is a constant. (b) Show how to solve for A H when K, is known a t two different temperatures. (c) Solve for in K, when AH = A H 0 aT are conbr2 CT%,where a, b, c, and stants. Given the following data for the viscosities, 7, of dilute solutions of ammonium chloride a t different concentrations, c [I.Am. Chem. SOG.,55, 636 (1933)],
41,500 1.987 X 573 =
n
n
-. n
-.
mtercept . = 1 log K. n = 1 n slope intercept, K = antilog slope (A V)+ 10. w = (A, + AB + Br)s' dw = dr (Ar f AB Br)a(AV)' - (AV)')lr [2(A +B)(Ar + AB (Ar AB Br)'
313
slope = 1
+
- (A,
+
( A V)'
+ AB + Br)=
-
+
++
2(A B) w , (Ar AB Br)
+
+ Br)
