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PHYSICAL CHEMISTRYB. Sc. 1st year

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PHYSICAL c h e m is t r yFor B.Sc. 1st Year

PAPER-II

(Strictly as per M.D.U., C.D.L.U. & K.U. Syllabus)

By

S. K IRA N VERMAN1Dept, o f Chemistry

Govt. College Chandigarh

VIVEK PATHANIADept, o f Chemistry Panjab University

Chandigarh

B. K. V ERM A N IDept, o f Chemistry

DAV College Chandigarh

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STH-8201-090-PHYSICAL CHEMISTRY I

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PREFACEThe Comprehensive Physical Chemistry (B. Sc. Part I) has been written strictly in accordance

with the new syllabus suggested by the UGC. It also contains other chapters to meet the requirements of the syllabi approved by the M.D.U., K.U. and the neighbouring universities in North India. Overall, it should suit the B.Sc. (1st year, Physical Chemistry) students of all the Indian universities. We have benefited greatly from the standard books on the subject written by Western and Indian authors. We make no claim for originality. Our humble contribution perhaps lies in the elucidation of mathematical derivations and basic concepts. Numerous solved examples have been scattered throughout the book. It goes without saying that Physical Chemistry has benefited from developments in modem physics and is a mathematical subject. It is mathematics which makes Chemistry both fascinating and frustrating! However, the mathematical demands here do not really extend beyond elementary differential and integral calculus and rudimentary algebra and coordinate geometry. All the necessary mathematical apparatus is elucidated by solved examples.

The chapter-end questions are limited to two categories, viz; review questions and multiple choice questions. Multiple choice questions have become the standard item in almost all the examinations and the entrance tests for admission to higher studies. It is important that the student has the necessary ability to tackle such questions.

We have benefited a great deal from discussions with several teachers and we take this opportunity to thank them here, especially Prof. D.S. Gill and Prof. R. P. Sharma, Department of Chemistry, Punjab University, Chandigarh, who contributed a great deal to this text. Also we wish to thank our inspiring publishers Sh. R. K. Gupta and his efficient staff for the unstinted help provided at all stages for publishing this book. We shall appreciate hearing from the student and the teaching community. Their feedback will go a long way in farther improvement of the subsequent editions of this book. We shall welcome constructive criticism from readers.

—Authors

SYLLABUS......................... . ...

(For Maharishi Dayanand, Rohtak and Kurukshetra Universities 2003-2004 and onwards)

PHYSICAL CHEM ISTRY (Theory) (PAPER-II, B.Sc. I YEAR)

Time allowed : 3 Hours] |M ax im um M arks : 37

Note : Ten questions will be set, two questions from each section. The candidate will be required to attempt five questions in all, selecting one question from each section. As far as possible questions will be short answer type and not essay type.

SECTIO N I

M athem atica l Concepts and C om pute rs 14 Hrs

Logarithmic relations, curve sketching, linear graphs and calculation o f slopes, differentiation o f functions like k , ev, x", sin x, “log x ; maxima and minima, partial differentiation and reciprocity relations. Integration of some useful/relevant functions ; permutations and combinations. Factorials. Probability.

General introduction to computers, different components o f a computer, hardware and soft­ware, input-output devices ; binary numbers and arithmetic.

SECTIO N II

Gaseous States 12 Hrs

Maxwell’s distribution of velocities and energies, (derivation excluded). Calculation o f root mean square, average and most probable velocities, mean free path including its derivation, collision diameter, collision number and frequency. Viscosity of gases, relationship between mean free path and coefficient o f viscosity, effect o f temperature and pressure on coefficient o f viscosity, calculation o f molecular diameter from coefficient o f viscosity. Degree o f freedom o f molecules, principle of equipartition o f energy.

van der Waals equation o f state, explanation of behaviour o f real gases by van der Waals equation. Application o f van der Waals equation in calculating Boyle’s temperature.

Critical Phenom ena : PV isotherms of real gases, continuity o f states, the isotherms o f van der Waals equation, relationship between critical constants and van der Waals constants, the law of corresponding states, liquefaction o f gases.

SECTION III

Liquid State 7 Hrs

Intermolecular forces, structure o f liquids : vacancy and free volume theory (a qualitative

description). Structural differences between solids, liquids and gases, (a qualitative description) properties o f liquids (surface tension, viscosity and refractive index) and their measurements.

Liquid crystals : Differences between liquid crystal, solid and liquid. Classification, structure o f nematic and cholestric phases. Thermography and seven segment cell.

Surface C h em is try 5 Hrs

Langmuir adsorption isotherm, BET equation (only elementary idea, derivation excluded) and its application. Gibbs adsorption equation and its applications.

Enzyme catalysis and mechanism of enzyme catalysis - Michaelis-Menten equation.

SEC TIO N IV

Solid State 10 Hrs

Definition o f space lattice, unit cell. Laws o f crystallography - (;') Law o f constancy of interfacial angles (ii) Law o f rationality o f indices (Hi) Law of symmetry. Symmetry elements in crystals.

X-ray diffraction by crystals. Derivation of Bragg equation. Determination o f crystal structure o f NaCl, KC1 and CsCl (Laue’s method and powder method).

SECTION V

Chemical Kinetics and Catalysis 12 Hrs

Chemical kinetics and its scope, rate o f a reaction, factors influencing the rate o f reaction - concentration, temperature, pressure, solvent, light, catalyst. Concentration dependence o f rates, mathematical characteristics o f simple chemical reactions - zero order, first order, second order, third order, pseudo unimolecular reactions, half life and mean life. Determination o f the order of reaction - differential method, method o f integration,' method o f half life period and isolation method.

Radioactive decay as a first order phenomenon.

Experimental methods o f chemical kinetics : conductometric, potentiometric, optical method, polarimetry and spectrophotometry.

Theories o f reaction rates - Effect o f temperature on rate o f reaction, Simple collision theory based on hard sphere model (mathematical derivation excluded), transition state theory (equilibrium hypothesis). Expression for the rate constant based on equilibrium constant and thermodynamic aspects.

CONTENTS

Chapters Pages

Syllabus (vz)

1. Mathematical Concepts and Computers 1

2. The Gaseous State 80

3. The Liquid State 132

4. Surface Chemistry 167

5. The Solid State 191

6. Chemical Kinetics and Catalysis 212

Appendix 257Bibliography 258University Questions 259

Mathematical Concepts and Computers(A) MATHEMATICAL CONCEPTS

L ogarithm ic R elations. Let fix) = e? , where x is a real variable, and e = 2.7182818....The algebraic operations with respect to exponential functions are

1e*. =ex+y ;e*/e> = ; — = ln x ...(1)

eThe inverse of e* is called the natural logarithm, In x, and e is called the base of the

logarithm. When e is replaced by 10 , then 1/10* = logx, called simply the common logarithm of x. The In x and log x are related by

In x = 2.303 log x ...(2)The fundamental relations satisfied by the natural (and also common) logarithms are

In xy = In x + In y ...(3)In (x/y) = In x - In y ...(4)

In xn - n In x ...(5)

Exam ple 1. Evaluate log3 243Solution : Let log3 243 = xHence, 3* = 243 = 3 x 3 x 3 x3 x 3 = 35

Thus, x = 5 ; so that log3 243 = 5

Exam ple 2. Find the logarithm of 5832 to the base 3^2.Solution : Let log3v(2 (5832) = x. Then, by definition,

(3V2f = 5832 = 8 x 729 = 23 x 36

= [(V2)2]3 x 36 = (3V2)6

Hence, x = 6, so that log3i/2 (5832) = 6.

x my nExam ple 3. Reduce to the simple form log t pzq

x my nSolution : loga = loga (xmy n) - loga (tpzq)

= loga (xm) + loga (yn) - loga (tp) - loga (zq) = m l o g x + n log y - p l o g t - q \ogn z

Exam ple 4. Show tha t logQ n - logft n x loga b Solution : Let x = log^ n and y = logt n

Then, ax - n and h* = n Hence, ax - which implies that

(<ax)Vy = b

log 6 = — = n ; so that log n - log n x log b y log b n

Corollary : If, in the above formula, we put n = a,Then, loga a = log6 a x logn b

or log6 a x logn b = 1 ( v loga a = 1)

C urve S k e tch in g , L in ea r G raphs an d S lo p esCurves and linear graphs play an important role in scientific research. Scientific data are

better appreciated, understood and analyzed if displayed as graphs. Mathematical functions and conic sections (circles, parabolas, ellipses and hyperbolas) can be better graphed with a graphing calculator. Linear graphs are described by the appropriate equation of a straight line. A linear equation in one variable has the form ax + b, where a # 0 and b are constants. The standard (general) form of the equation of a straight line is

ax + by + c - 0 ...(6)where not both a and b are zero, and a, b, c are real numbers. The slope, m, of a line is the ratio of the change in y compared with the change in x\ thus,

change in y _ dy change in x dx

Where dy is infinitesimal change in y corresponding to infinitesimal change in x. If CCj^) and (x2y 2,) are two points on a line and m is the slope of the line, then

m =

m - y 2 ~ y \ when x2 * x y .(8 )2 1

The angle which a line makes with the positve direction of x-axis measured in the anti­clockwise direction is called the inclination (or angle of inclination), denoted by 6 (Fig. 1)

The following facts should be borne in mind :(a) The inclination of a line parallel to x-axis or the x-axis itself is 0°. ib) The inclination of a line parallel toy-axis ory-axis itself is 90°.If 9* 90° is the inclination of a line, then the slope, m, of the line is defined asm = tan 9. Note th a t since tan 0 is not defined when 9= 90°, the slope of a vertical line is not defined. Exam ple 5. Calculate the slope of the line passing through the points (6, -8) and (5, 2).

y2- y 1 2 - ( - 8)Solution : x — x. 5 - 62 1

= - 10

The slope of the line is negative. This means that, as we view the graph of the line from left to right, as x increases y decreases.

P ara lle l and P e rp e n d ic u la r L ines : Two non-vertical lines are parallel if and only if the slopes of the lines are equal, i.e., m l = m2. Two non-vertical lines are perpendicular if and only if the product of their slopes is - 1 , i.e., m ]m2 = - 1.

Exam ple 6. Determine whether the line containing the points A and B is parallel, per­pendicular, or neither to the line containing the points C and D.

(a) A (2, 4), B (3, 8), C (5, 1), and D (4, - 3)(b)A (2, - 3), B (- 4, 5), C (0, - 1), and D (- 4, - 4)(c) A (1, 9), B (4, 0), C (0, 6), and D (5, 3)Solu tion:

__ 8 - 4 , __ - 3 - 1( а )Slope (AB) = m l = = 4 ; slope (CD) = m 2 = 4 _ g = 4

Since m 1 = m 2, hence lines AB and CD are parallel.— 5 - ( - 3 ) 4 , — - 4 — (— 1) 3

(б)Slope (AB) = m 1= _ 4 _ 2 = ~ J ’> sloPe (CD) = m 2 = _ 4 _ Q = 4

Since m lm 2 = (- 4/3) (3/4) = - 1, the lines AB and CD are perpendicular.__ 0 - 9 __ 3 - 6 3

(c) Slope (AB) = m 1= = - 3 ; slope (CD) = m 2 = g _ Q = - g

Since m 1 * m 2 and also m lm2 * - 1, hence the lines AB and CD are neither parallel nor perpendicular.

Exam ple 7. Determine whether the given three points are collinear or not:(a) (0, 3), (1, 1) and (2, - 1) ; (b) (1, 5), (- 2, - 1) and (- 3, - 4).Solu tion:

1 - 3 - 1 - 1(а)m i = = _ 2 ; t t t = _ 2Since the line between (0, 3) and (1, 1) and the line between (1 , 1) and (2, -1) have the

same slope, the given points are collinear.- 1 - 5 — 4 — (— 1)

(б)™ i= r y r r = 2 ; _ 3 _ (_ 2y = 3

Since the slope of the line between (1, 5) and ( - 2 , - 1 ) and the slope of the line between (- 2, -1) and (-3, - 4) are different, the given points are not collinear.

S lope-In tercep t Form of th e E quation of a Line

If a line has slope m and y-intercept (0, 6), then for any point Oc, y), where x * 0 on the line, y - bm = , and y = mx + bx - 0

Thus, the slope-intercept form of the equation of a line with slope m and irtercept b is y = mx + b ...(9)

Exam ple 8. Write down the equation of the line with slope m = -2/3 and y-intercept 6 = 6. Solution : y = mx + 6 = (- 2/3) x + 6or 2x + 3y = 18Slope-Point Form of th e E quation of a LineIf a line has slope m and passes through a point (Xj.y,), then for the point (x, y) on the line,

we have m = (y - y ^ K x - x J and (y - y x) = m (x -Xj). Thus, the slope-point form of the equation of the line is

( y - y t) = m (x-X j) ...(10)In te rc ep t Form of th e E quation of th e L ine

If a line has x intercept a and y intercept 6, it goes through the points (a, 0) and (0, 6). The equation of the line is

i 0 - b y - b = - (x 0), if a * 0a - 0

which simplifies to 6x + ay = ab. If both a and 6 are non-zero, we obtainx y a 6 = 1 ...(11)

Which is the desired equation.Some of the commonly encountered graphs in elementary physics involve (a) Displace­

ment-time graphs, and (6) velocity-time graphs. The shortest distance travelled by a moving object between the initial and final position is called displacement, S and th£ time-rate of change of S is called the velocity: v= dS/dt . Typical displacement-time graphs are shown in Fig. 2.

(c) (d)Fig. 2. The displacement-time graphs.Fig. 2 (a) shows positive uniform velocity since displacement is increasing with time,

whereas Fig. 2 (b) shows negative uniform velocity since displacement is decreasing with time. Fig. 2 (c) shows positive variable velocity since the displacement is not increasing uniformly with time, whereas Fig. 2 (d ) shows negative variable velocity since the displacement is not decreasing uniformly with time.

Typical velocity-time graphs are shown in Fig. 3. Fig. 3 (a) shows tha t the moving object has a uniform positive acceleration because the velocity is increasing with time. Recall thatacceleration is defined as the time-rate of change of velocity : a =dv/dt . Fig. 3 (b) shows that the object is moving with a uniform negative acceleration (retardation) because the velocity is decreasing with time. Again, Fig. 3 (c) shows variable acceleration, whereas Fig. 3 (d) shows variable retardation.

(a) (b )

(c) (d)

Exam ple 9. Draw the graph of the linear equation y = x + 2.Solution : The given equation is of first degree in the variables x and y ; thus, it is the

equation of a straight line. We draw up the following value table :

X 0 1 2

y 2 3 4

which gives three representative points A (0, 2), B (1, 3) and C (2, 4). The graph is shown on the Cartesian axis system in Fig. 4.

Similarly, we can solve two simultaneous linear equations in two variables by plotting their graphs. The point of intersection of .''e two lines gives the solution of the given equa­tions.

D ifferen tiation . The derivative of a function y = f(x) at a point x is defined as

“I f ) - %where Ax = h and Ay = f i x + h) - f i x ) are the increments in the variables x and y, respectively. The derivative of f ix) is denoted by dy/dx, provided the limit exists, i.e.,

dy f i x + h) - fix)dx h .(12)

and is interpreted as the rate of change of y w.r.t. x. The process of finding a derivative is called differentiation. From the geometrical point of view, the derivative of a function f ix) at a given point represents the slope of the tangent drawn to the curve of y = f ix) at the point where the function is defined.

Exam ple 10. Find the derivative of y = x2 + 3x + 5Solu tion: y + Ay = (x + h)2+ 3 {x + h) + 5

= x2 + 2xh + 3x + 3h + 5

Ay (2x + 3)h + h ‘ Ax ' hdy

= 2x + 3 + h

, = limd x Ax 0 Ax /i -> 0

Ay = lim (2* + 3 + h) = 2* + 3.

Exam ple 11. Show tha t sin x = cos x

c , d .. sin(x + /i) - sinx Solu tion : -r~ sin x = lim ----------- ------------dx u n hh -> 0

/ h. . h2 cos u + 77) sin ~x= lim __________A_____4l

0

= limh—>0

cos2 /

h

sin (h / 2) h i 2

v s in (/i/2 ) , lim — — = 1h-+ 0 h / 2

Exam ple 12. Defferentiate xn with respect to x from first principles.

Q . dxn (x + h)n - x nSolution : —— = limdx h ->0

= lim h -» 0

(x + h)n n

(x + h ) - x

x n - anWe know that lim ---------- = nan~l ; hence,

x - a

(x + h)n - x n ,lim —— —r;-------= nxn ; thus,h - > 0

dxndx

(x + h ) - x

= nxn 1

Exam ple 13. Differentiate e* with respective to x from first principles.

dexSolution = lim„x+he - e

dx h : ; 0 h

limh ->0

ex(eh - 1) - e* limh ->0

eh - l

v eh~ 1 1 .• lim — -— = ljh ->0 h

Comprehensive Physical Chemistry

Publisher : Laxmi Publications ISBN : 9788170086901Author : B K Vermani, SKiran Vermani & VivekPathania

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