Chemistry MCQs for NEET — Practice Questions with Answers

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For isothermal expanison in case of an ideal gas:-

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Explanation

For ideal gas isothermal expansion

H=0G=-TS

At 25°C, for the process H2O(l)H2O(g);

G° is 8.6 kJ. The vapour pressure of water at this temperature is nearly:-

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Explanation

At equilibrium G° = -RTlnKeq

 KeqPH2O/P° and P°= 1 bar

What will be the standard enthalpy of reaction for the following reaction using the listed enthalpies of reaction:-

3Co(s) + 2O2(g)Co3O4(s);       H=?

2Co(s) + O2(g) 2CoO(s)          H1= -475.8 KJ

6CoO(s) + O2(g) 2Co3O4(s)     H2 = -355.0 KJ

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Explanation

H°=32H1° + H2°2

       = -713.7 - 177.5 = -891.2 KJ

For given following equations and H° values determine the enthalpy of reaction at 298 K for the reaction 

C2H4(g) + 6F2(g)  2CF4(g) + 4HF(g)

H2(g) + F2(g)  2HF(g)       H1°= -537 kJ

C(s) + 2F2(g) CF4(g)         H2°=-680 kJ

2C(s) + 2H2(g) C2H4(g)    H3°= 52 kJ

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Explanation

 H°=2H1° + 2xH2°-H3°

If Hf°(C2H4) and Hf°(C2H6) are x1 and x2 kcal/mol then heat of hydrogenation of C2H4 is :-

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Explanation

3. Target reaction C2H4 + H2 C2H6

Given reaction

(i) 2C + 2H2 C2H4    H1=x1

(ii) 2C + 3H2C2H6    H2= x2

(ii)-(i) = x2 - x1

For oxidation of iron, 4Fe(s) + 3O2(g) 2Fe2O3(s), Hr°=-1648x103 J mol-1 entropy change is:

-549.4 J k-1mol-1 at 298 K:-

The reaction is

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Explanation

ST=Ssystem+SSurr.       = + ve

18 g of ice is converted into water at 0°C and 1 atm. The entropy of H2O(s) and H2O(l) are 38.2 and 60 J K-1 mol-1 respectively. the enthalpy for this conversion will be

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Explanation

 Sr= 60-38.2=21.8 JK-1 mol-1

        ..H=21.8 x 273 = 5951.4 J/mol

For a given reaction, H =35.5 kJmol-1 and S = 83.6JK-1 mol-1. The reaction is spontaneous at: (Assume that H and S do not vary with temperature)

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Explanation

(b) According to Gibbs-Helmholtz equation, Gibbs energy (G) = H - TS

where, H = Enthalpy change

S = Entropy change

T = Temperature

For a reaction to be spontaneous G <0.

Gibbs -Helmholtz equation becomes.

G = H -TS<0

or, H < TS
T > H / S

=35.5 KJmol-1/ 83.6 JK-1mol-1

=35.5 x 1000J mol-1 / 83.6JK-1mol-1

=425 K

T>425K

A gas is allowed to expand in a well insulated container against a constant external pressure of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L. The change in internal energy U of the gas in joules will be

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Explanation

(c) Key concept According to first law of thermodynamics,

U = q + w

where, U =internal energy

q = heat absorbed or evolved, w = work done.

Also, work done against constant external pressure (irreversible process).

W = -Pext V.

Work done in irreversible process,

w = -Pext V = - pext (V2 - V1)

= -2.5 atm (4.5 L - 2.5 L)

= - 5 L atm = - 5 x 101.3 J

= - 505 J

Since, the system is well insulated, q =0

U = w = - 505 J

Hence, change in internal energy, U of the gas is - 505 J.

 

For a sample of perfect gas when its pressure is changed isothermally from pi to pf, the entropy change is given by

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Explanation

(b) Entropy change is given as,

     S = nCpln(Tf/Ti) + nRln(pi/pf)           ....(i)

For isothermal process, Ti = Tf

...  nCpln(Tf/Ti) =0 [ln1 = 0]

From Eq (i)  S = nRln(pi/pf)

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