Chemical Thermodynamics MCQs for NEET — Chemistry Questions with Answers

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In which of the following reactions does the heat change represent the heat of formation of water [EAMCET 1991]

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Explanation

Heat of formation is the formation of one mole of the substance from its elements.

Based on the following thermochemical equations

H2O(g)+C(s)CO(g)+H2(g);ΔH=131kJ

CO(g)+12O2(g)CO2(g);ΔH=282kJ

H2(g)+12O2(g)H2O(g);ΔH=242kJ

C(s)+O2(g)CO2(g);ΔH=XkJ

The value of X is [CBSE PMT 1992]

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Explanation

eq. (i) + eq. (ii) + eq. (iii) gives

X=131282242=393kJ.

If enthalpies of formation of C2H4(g), CO2(g) and H2O(l) at 25°C and 1 atm pressure be 52, – 394 and –286 kJ mol–1 respectively, the enthalpy of combustion of C2H4(g) will be [CBSE PMT 1995; AIIMS 1998; Pb. PMT 1999]

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Explanation

C2H4+3O22CO2+2H2O

ΔHreaction=[2×ΔHfo(CO2)+2×ΔHfo(H2O)] [ΔHfo(C2H4)+3  ×ΔHfo(O2)]

=[2(394)+2(286)][52+0]=1412kJ.

Ozone is prepared by passing silent electric discharge through oxygen. In this reaction [AFMC 1998]

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Explanation

3O22O3 – energy is given out. 

Combustion of glucose takes place according to the equation, C6H12O6+6O26CO2+6H2O, ΔH=72kcal. How much energy will be required for the production of 1.6 g of glucose (Molecular mass of glucose = 180 g) [AFMC 1999]

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Explanation

ΔHper1.6g=72×1.6180=0.64kcal.  

Which of the following compounds will absorb the maximum quantity of heat when dissolved in the same amount of water ? The heats of solution of these compounds at 25°C in kJ/mole of each solute is given in brackets [AMU (Engg.) 2000]

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Explanation

More +ve is ΔHs more is heat of solution.

A system is changed from state A to state B by one path and from B to A another path. If E1 and E2 are the corresponding changes in internal energy, then [Pb. PMT 2001]

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Explanation

ΔE = 0 for a cyclic process.

If (i) C+O2CO2, (ii) C+1/2O2CO, (iii) CO+1/2O2CO2, the heats of reaction are Q, –12, –10 respectively. Then Q = [Orissa JEE 2004]

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Explanation

C+O2CO2; ΔH = q

C+1/2O2CO; ΔH = –12 …..(i)

CO+1/2O2CO2; ΔH = –10 …..(ii)

adding equation (i) and (ii) we can get

ΔH = –12 + (–10) = –22

Adsorption of gases on solid surface is generally exothermic because [IIT JEE (Screening) 2004]

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Explanation

Due to randomness of particles is reduced since entropy decreases.

Two mole of an ideal gas is expanded isothermally and reversibly from 1 litre ot 10 litre at 300 K. The enthalpy change (in kJ) for the process is [IIT JEE (Screening) 2004]

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Explanation

ΔH = nCp ΔT

The process is isothermal therefore

ΔG = 0; ∴ ΔH = 0

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