Chemistry MCQs for NEET — Practice Questions with Answers

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The Kirchhoff's equation gives the effect of....on heat of reaction.

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Explanation

Kirchhoff's equation is:

∆H2-∆H1=∆CPT2-T1

 

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The temperature of 5 ml of a strong acid increases by 5∘ when 5 ml of a strong base is added to it. If 10 ml of each is mixed, temperature should increase by-

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Explanation

When 5 ml of strong acid reacting with 5 ml of strong base net, heat evolved = ∆H

∴ temperature increase = ∆Hms

when 10 ml of each are reaction, the heat evolved = 2∆H;

but mass of the solution = 2m

∴ temperature increase = 2∆H2ms=∆Hms

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The heat of neutralisation of HCl by NaOH is -55.9 kJ/mole. If the heat of neutralisation of HCN by NaOH is -12.1 kJ/mole, then energy of dissociation of HCN is-

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Explanation

The heat of neutralisation = -55.9 kJ

∴ The heat of neutralisation of HCN by NaOH = the heat of neutralisation + the energy of dissociation of HCN

= -12.1 kJ

or the energy of dissociation of HCN

= +55.9-12.1 = 43.8 kJ

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For a reaction at equilibrium-

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Explanation

For a reaction at equilibrium ∆G=0 and not ∆G∘.

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Using only the following data:

(I) Fe2O3s+3COg⇌2Fe(s)+3CO2(g);∆H∘=-26.8 kJ

(II) Fes+COg⇌FeOs+COg;∆H∘=+16.5 kJ

the ∆H∘ value, in kilojoules, for the reaction

Fe2O3s+COg→2FeOs+CO2g is calculated to be:

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Explanation

from equation (I) + (2×II) ;     ∆H∘= 6.2 kJ

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82 litres of carbon dioxide are produced at a pressure of 1 atm by the action of acid on a metal carbonate. The work done by the gas (in calories) in pushing back the atmosphere is (R = 0.082 litre-atm deg-1mol-1)

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Explanation

The work done by a gas is given by

W= P∆V

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Among the following, the reaction for which ∆H=∆E is-

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Explanation

When ∆n=0, ∆H=∆E  because ∆H=∆E+RT ∆n.

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For which change ∆H≠∆E

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Explanation

∆H=∆E+∆nRT

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Enthalpy change when 1.00 g water is frozen at 0∘C, is :

∆Hfus=-1.435 kcal mol-1

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Explanation

∆H(per g) = -1.43518 kcal =  - 0.0797 kcal g

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One mole of hydrogen gas at 25∘C and 1 atm pressure is heated at constant pressure until its volume has doubled. Given that Cv for hydrogen is 3.0 cal deg-1 mol-1, the ∆H and ∆E for this process are-

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Explanation

∆H=CP∆T and ∆E=CV∆T.

CP-CV=R using these equations after calculating the final temperature, the desired result is obtained.

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