Chemistry MCQs for NEET — Practice Questions with Answers

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Enthalpy of the reaction,

 CH4(g) + 1/2 O2(g) CH3OH (l), is negative. If enthalpy of combustion of CH4 and CH3OH are x and y respectively, then which relation is correct?                                 

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Explanation

(b) CH4(g) + 1/2 O2(g) CH3OH (l)

      ..H=[(Hcof CH3OH)-(Hc of CH4)]

                = -[(-y)-(-x)]

                =-[y+x] =y-x

       ... x<y

Alternate explanation: 

CH4+2O2CO2+2H2O            H=xCH3OH+32O2CO2+2H2O    H=y

- (2)

CH4+12O2CH3OH x-y

As x - y < 0

x < y

In the reaction, H and S both are positive. In which of the following cases, the reaction would not be spontaneous?

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Explanation

(d) G=+ ve in a, and ZERO in b,c

When an ideal gas is compressed adiabatically and reversibly, the final temperature is:

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Explanation

(a) q=U-W, if adiabatic process q=0, then -U = -W, i.e, a decrease in free energy brings in work done by the system (-W).

The entropy change in the fusion of one mole of a solid melting at 27°C (latent heat of fusion is 2930 J mol-1) is                                                                                              

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Explanation

(a) Entropy, Sf=Hf/Tf = Fusion enthalpy/Temperature

                  Sf = 2930 J mol-1/300 K = 9.77 JK-1mol-1

The maximum work done in expanding 16 g oxygen at 300 K and occupying a volume of 5 dm3 isothermally until the volume becomes 25 dm3 is:

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Explanation

(a) -W = +2.303nRTlogV2V1

      -W = 2.303 x 16/32 x 300 x 8.314log(25/5)

      -W=2.01 x 103 J

1 mole of an ideal gas at 25°C is subjected to expand reversibly ten times of its initial volume. The change in entropy of expansion is:

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Explanation

(a) S=q/T =2.303nRTTlogV2V1=2.303 x 1 x 8.314log10

                                          = 19.15 JK-1mol-1

 During an adiabatic process:

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Explanation

(c) An adiabatic process is one in which exchange of heat is not taking place in between system and surroundings. This can be made by putting insulation at the boundaries of system.

Heat of combustion H° for C(s), H2(g) and CH4(g) are -94, -68 and -213 kcal/mol. Then, H° for 

C(s) + 2H2(g) CH4(g) is                                                                 

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Explanation

(a) For reaction,

C(s) + 2H2(g) CH4(g) , H°=?

 H°=-[Hc° of CH4) -(Hc°of C +2x Hc° of H2)]

              C+ OCO2,    H° = -94 kcal   ...(i)

           2H2 + O2 2H2O,  H°= -68 x 2 kcal         ...(ii)

     CH4 + 2O2 CO2 + 2H2O,  H°=-213 kcal    .....(iii)

On adding eqs. (i) and(ii) and then subtracting eq (iii)

              = -[(-213)-(-94+2x-68)] kcal/mol

              = -[-213 + 230] = -17 kcal/mol

 

If 50 calorie are added to a system and system does work of 30 calorie on surroundings, the change in internal energy of system is:

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Explanation

(a) q=U + W, its work done by the system.

    50=U + 30

 ...   U = 20 cal

 

The internal energy change when a system goes from state A to B is 40 kJ/mol. If the system goes from A to B by a reversible path and returns to state A by an irreversible path. What would be the change in internal energy?

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Explanation

(d) In a cyclic process, U=0

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