NEET UG CHEMISTRY MCQ: 14 TH GROUP ELEMENTS (P-BLOCK)

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Application +4 | -1

The enthalpies of combustion of

\mathrm{C}

(graphite) and

\mathrm{C}

(diamond) are

-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}

and

-395.3 \mathrm{~kJ} \mathrm{~mol}^{-1}

, respectively. The enthalpy change of the reaction

\mathrm{C}

(graphite)

\rightarrow

\mathrm{C}

(diamond) is

\mathrm{P} \times 10^2 \mathrm{~J} \mathrm{~mol}^{-1}

. Then

\mathrm{P}

is...

View answer and explanation

Correct option: A

Option A (18) is correct because the enthalpy change for the conversion of graphite to diamond is the difference between their enthalpies of combustion, which equals 1.8 kJ/mol or 1800 J/mol, hence P = 18.

NEET UG CHEMISTRY MCQ: 14 TH GROUP ELEMENTS (P-BLOCK)

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