Chemical Thermodynamics MCQs for NEET — Chemistry Questions with Answers

The combustion of cyclopropane involves the formation of CO2 and H2O from the cyclopropane molecule. The enthalpy change can be calculated using the given enthalpies of formation and the enthalpy of isomerization of cyclopropane to propene, which acts as an intermediate step.

As we know that

$∆$H = $∆$E + P$∆$V

or $∆$H = $∆$E + $∆$nRT

where $∆$H $\to$ change in enthalpy of system (standard heat at constant pressure)

$∆$E $\to$ Change in internal energy of system (Standard heat at constant volume)

An $\to$ no. of gaseous moles of product - no. of gaseous moles of reactant

R $\to$ gas constant

T $\to$ absolute temperature

If $∆$n = 0 for reactions which is carried out in an open container, therefore $∆$H = $∆$E

So for reaction (1) $∆$n = 2 - 2 = 0

Hence, for reaction (1), $∆$H = $∆$E

Decomposition of H2O2 is a spontaneous process hence Delta G value is negatively hence decrease in Free energy will occur.... 

(A) Randomness increases on changing the phase from soilid to liquid or from liquid to gas.

According to the provided text, 'The entropy of any pure crystalline substance approaches zero as the temperature approaches absolute zero. This is called third law of thermodynamics.' The other options describe the Second Law, First Law, and a consequence of the Second Law, respectively.

The context states, 'This is so because there is perfect order in a crystal at absolute zero.' While molecular motion significantly decreases, the fundamental reason for zero entropy in a perfect crystal at 0 K is the perfect order, meaning only one microstate is accessible.

The NCERT text explicitly states, 'The statement is confined to pure crystalline solids because theoretical arguments and practical evidences have shown that entropy of solutions and super cooled liquids is not zero at 0 K.'

The text highlights, 'The importance of the third law lies in the fact that it permits the calculation of absolute values of entropy of pure substance from thermal data alone.'

The context explains, 'Mol ecules of a substance may move in a straight line in any direction, they may spin like a top and the bonds in the molecules may stretch and compress. These motions of the molecule are called translational, rotational and vibrational motion respectively.' Spinning like a top corresponds to rotational motion.

The text states, 'When temperature of the system rises, these motions become more vigorous and entropy increases.'