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 Multiple Choice QuestionsMultiple Choice Questions

1.

The standard enthalpy of formation of NH3 is– 46.0 kJmol–1. If the enthalpy of formation of H2 from its atoms is – 436 kJ mol–1 and that of N2 is – 712 kJ mol–1,the average bond enthalpy of N – H bond is NH3 is

  • -964 kJ mol-1

  • +352 kJ mol-1

  • +1056 kJ mol-1

  • -1102 kJ mol-1


B.

+352 kJ mol-1

NH subscript 3 space left parenthesis straight g right parenthesis space rightwards arrow with space on top space 1 half space straight N subscript 2 space left parenthesis straight g right parenthesis space space plus space 3 over 2 space straight H subscript 2 space left parenthesis straight g right parenthesis
increment straight H degree space equals space minus space increment minus straight H degree subscript straight f space left parenthesis NH subscript 3 right parenthesis space equals space minus space left parenthesis negative 46 right parenthesis space equals space 46 space kJ space mol to the power of negative 1 end exponent
Also comma space increment straight H degree space equals space 3 increment straight H subscript straight N minus straight H end subscript space plus 1 half increment straight H subscript straight N identical to straight N end subscript space plus 3 over 2 space increment straight H subscript straight H minus straight H end subscript
space 46 space equals space 3 space increment straight H subscript straight N minus straight H end subscript space plus 1 half space left parenthesis negative 712 right parenthesis space plus space 3 over 2 space left parenthesis negative 436 right parenthesis
increment straight H subscript straight N minus straight H end subscript space equals space 1 third left square bracket 1056 right square bracket space equals space plus 352 space kJ space mol to the power of negative 1 end exponent
670 Views

2.

The heats of combustion of carbon and carbon monoxide are −393.5 and −283.5 kJ mol−1, respectively. The heat of formation (in kJ) of carbon monoxide per mole is:

  • 676.5

  • -676.5

  • -110.5

  • 110 s


C.

-110.5

C(s) + O2 (g) → CO2 (g); ΔH = -393.5 kJ mol-1 ... (i)

CO + O2/2 → CO2 (g); ΔH = - 283.5 kJ mol-1 ....(ii)
On subtracting Eq. (ii) from Eq. (i), we get

C (s) + O2/2 (g) → CO (g)


ΔH = (-393.5 + 283.5) kJ mol-1 = - -110 kJ mol-1

593 Views

3.

The entropy change involved in the isothermal reversible expansion of 2 moles of an ideal gas from a volume of 10 dm3 to a volume of 100 dm3 at 27°C is

  • 38.3 J mol-1 K-1

  • 35.8 J  mol-1 K-1

  • 32.3 J  mol-1 K-1

  • 42.3 J mol-1 K-1


A.

38.3 J mol-1 K-1

increment straight S space equals space nR space In space straight V subscript 2 over straight V subscript 1
equals space 2.303 space nR space log space straight V subscript 2 over straight V subscript 1
space equals space 2.303 space straight x space 2 space straight x space 8.314 space straight x space log space 100 over 10
equals space 38.3 space straight J space mol to the power of negative 1 end exponent straight k to the power of negative 1 end exponent
789 Views

4.

The standard Gibbs energy change at 300 K for the reaction, 2A  ⇌ B +C is 2494.2J at a given time, the composition of the reaction mixture isleft square bracket straight A right square bracket space equals space 1 half comma space left square bracket straight B right square bracket space equals 2 and left square bracket straight C right square bracket space equals space 1 half, The reaction proceeds in the [R= 8.314 JK/mol, e = 2.718]

  • forward direction because Q>Kc

  • reverse direction because Q>Kc

  • forward direction because Q < Kc

  • reverse direction because Q < Kc


B.

reverse direction because Q>Kc

We know,
ΔG = ΔGo + RTlnQ .. (i) 
Given,
ΔGo  = 2494.2J
 straight Q space equals fraction numerator left square bracket straight B right square bracket right square bracket straight C right square bracket over denominator left square bracket straight A right square bracket squared end fraction space equals space fraction numerator 2 space straight x begin display style 1 half end style over denominator open parentheses begin display style 1 half end style close parentheses squared end fraction space equals space 4
thus,
putting the value in equation (i)
 = 2494.2 +8.314 + 300 In 4
= 28747.27 J
= positive value
Also, we have
increment straight G space equals space RT space ln space straight Q over straight K
If ΔG is positive, Q >K
therefore, reaction shifts in the reverse direction

600 Views

5.

For the complete combustion of ethanol, C2H5OH (l) + 3O2 (g) → 2CO2 (g) + 3H2(l), the amount of heat produced as measured in a bomb calorimeter, is 1364.47 kJ mol-1 at 25oC. Assuming ideality the enthalpy of combustion, CH, for the reaction will be (R = 8.314 JK-1 mol-1)

  • -1366.95 kJ mol-1

  • -1361.95 kJ mol-1

  • -1460.50 kJ mol-1

  • -1350.50 kJ mol-1


A.

-1366.95 kJ mol-1

C2H5OH (l) + 3O2 (g) → 2CO2 (g) + 3H2(l),
∆U = - 1364.47 kJ/mol
∆H = ∆U +∆ngRT
∆ng = -1
∆H = - 1364.47 +fraction numerator negative 1 space straight x space 8.314 space straight x space 298 over denominator 1000 end fraction
[Here, value of R in unit of J must be converted into kJ]
 = - 1364.47-2.4776
 = -1366.94 kJ/mol
980 Views

6.

The energy required to break one mole of Cl— Cl bonds in Cl2 is 242 kJ mol. The longest wavelength of light capable of breaking a single Cl — Cl bond is
(c= 3 x 108 ms–1and NA = 6.02 x 1023 mol–1)

  • 594 nm

  • 640 nm

  • 700 nm

  • 494


D.

494

Energy, E = NA
hv space equals space straight N subscript straight A space straight x space hc over straight lambda
or space straight lambda space equals space straight N subscript straight A space straight x space hc over straight E
straight lambda space equals space fraction numerator 6.626 space straight x space 10 to the power of negative 34 end exponent space straight x space 3 space straight x space 10 to the power of 8 space straight x space 6.02 space space straight x space 10 to the power of 23 over denominator 242 space straight x space 10 cubed end fraction
space 494 space straight x space 10 to the power of negative 9 end exponent space straight m equals space 494 space nm

232 Views

7.

The following reaction is performed at 298 K
2NO(g) + O2 (g) ⇌ 2NO2 (g)
The standard free energy of formation of NO (g) is 86.6 kJ/mol at 298 K. What is the standard free energy of formation of NO2 (g) at 298 K? (KP = 1.6 x 1012)

  • R (298) In (1.6 x 1012)-86600

  • 86600 + R (298) In (1.6 x 1012)

  • 86600 - In(1.6 x 1012)/R(298)

  • 0.5[2 x 86600-R(298)In (1.6 x 1012)]


D.

0.5[2 x 86600-R(298)In (1.6 x 1012)]

For the given reaction,
2NO(g) + O2 (g) ⇌ 2NO2 (g)
Given ,

increment straight G subscript straight f superscript straight o left parenthesis NO right parenthesis space equals space 86.6 space kJ divided by mol
increment straight G subscript straight f superscript straight o left parenthesis NO subscript 2 right parenthesis space equals ?
straight K subscript straight p space equals space 1.6 space straight x space 10 to the power of 12
Now, we have,

increment straight G subscript straight f superscript degree space equals space 2 increment straight G subscript straight f left parenthesis NO subscript 2 right parenthesis end subscript superscript straight o space minus space left square bracket 2 increment straight G subscript straight f left parenthesis NO right parenthesis end subscript superscript straight o space plus space increment straight G subscript straight f left parenthesis straight O subscript 2 right parenthesis end subscript superscript straight o right square bracket
equals negative RT space In space straight K subscript straight p space equals space 2 increment straight G subscript straight f superscript straight o subscript left parenthesis NO subscript 2 right parenthesis end subscript space minus left square bracket 2 space straight x 86600 space plus 0 right square bracket
increment straight G subscript straight f superscript straight o subscript left parenthesis NO subscript 2 right parenthesis end subscript space equals space 1 half left square bracket space 2 space straight x space 86600 space minus straight R space straight x space 298 space In space left parenthesis 1.6 space straight x space 10 to the power of 12 right parenthesis right square bracket
increment straight G subscript straight f superscript straight o subscript left parenthesis NO subscript 2 right parenthesis end subscript space equals space 0.5 space left square bracket space space 2 space straight x space 86600 minus space straight R space straight x space left parenthesis 298 right parenthesis In space left parenthesis 1.6 space straight x space 10 to the power of 12 right parenthesis right square bracket

564 Views

8.

The standard reduction potentials for Zn2+/ Zn, Ni2+/ Ni, and F2+/ Fe are –0.76, –0.23 and –0.44 V respectively. The reaction X + Y2+ → X 2+ + Y will be spontaneous when

  • X = Ni, Y = Fe

  • X = Ni, Y = Zn

  • X =Fe, Y= Zn

  • X = Zn, Y = Ni


D.

X = Zn, Y = Ni

X = Zn, Y = Ni
Zn + Ni2+  →Zn2+ + Ni

straight E subscript cell superscript straight o space equals space straight E subscript zn divided by zn to the power of 2 plus end exponent end subscript superscript straight o space plus space straight E subscript Ni to the power of 2 plus end exponent divided by Ni end subscript superscript straight o
space equals space 0.76 minus 0.23
space equals space 0.53 space straight V
straight E subscript cell superscript 0 space greater than thin space 0

206 Views

9.

A piston filled with 0.04 mol of an ideal gas expands reversibly from 50.0 mL to 375 mL at a constant temperature of 37.00C. As it does so, it absorbs 208J of heat. The values of q and w for the process will be:(R = 8.314 J/mol K) ( ln 7.5 = 2.01)

  • q =+208J, W = - 208 J

  • q =-208 J, W =-208 J

  • q=-208J, W = +208 J

  • q =+208 J, W = +208 J


A.

q =+208J, W = - 208 J

From first law of thermodynamics, ΔE = q+W for an isothermal expansion. 
Hence, q =-W
q= +208 J
W =-208 J [expansion work]

541 Views

10.

The incorrect expression among the following is

  • fraction numerator increment space straight G subscript system over denominator increment straight G subscript total end fraction space equals space minus straight T
  • In isothermal process

    straight W subscript reversible space equals negative nRT space ln straight V subscript straight f over straight V subscript straight i

  • In space straight K space equals space fraction numerator increment straight H to the power of straight o minus straight T increment straight S over denominator RT end fraction
  • straight K space equals space straight e to the power of increment straight G to the power of straight o divided by RT end exponent

C.

In space straight K space equals space fraction numerator increment straight H to the power of straight o minus straight T increment straight S over denominator RT end fraction

Option C has incorrect expression. The correct expression is,

increment straight G to the power of straight o space equals space increment straight H to the power of straight o space minus straight T increment straight S to the power of straight o
increment straight G to the power of straight o space equals nRT space log space straight K
therefore equals negative RT space log space straight K space equals increment straight H to the power of straight o minus straight T increment straight S to the power of straight o
therefore space log space straight K space equals space minus space open parentheses fraction numerator increment straight H to the power of straight o minus straight T increment straight S to the power of straight o over denominator RT end fraction close parentheses

334 Views