${C_{{\rm{diamond}}}} + {{\rm{O}}_2}\left( {\rm{g}} \right) \to {\rm{C}}{{\rm{O}}_2}\left( {\rm{g}} \right);\Delta H = - 395{\rm{kJ}} \ldots \ldots {\rm{.}}\left( {\rm{i}} \right)$ ${C_{{\rm{graphite}}}} + {{\rm{O}}_2}\left( {\rm{g}} \right) \to {\rm{C}}{{\rm{O}}_2}\left( {\rm{g}} \right);\Delta H = - 393.5{\rm{kJ}} \ldots \ldots {\rm{.}}\left( {{\rm{ii}}} \right)$ The ∆H, when diamond is formed from graphite, is :

JEE CHEMISTRY MCQ: THERMODYNAMICS

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Application +4 | -1

{C_{\mathrm{diamond}}} + {\mathrm{O}_2}\left( \mathrm{g} \right) \to \mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right);\Delta H = - 395\mathrm{kJ} \ldots \ldots \mathrm{.}\left( \mathrm{i} \right)

{C_{\mathrm{graphite}}} + {\mathrm{O}_2}\left( \mathrm{g} \right) \to \mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right);\Delta H = - 393.5\mathrm{kJ} \ldots \ldots \mathrm{.}\left( {\mathrm{ii}} \right)

The ∆H, when diamond is formed from graphite, is :

View answer and explanation

Correct option: B

+ 1.5\mathrm{kJ}

Option B (+1.5 kJ) is correct because the enthalpy change for the conversion of graphite to diamond is calculated by subtracting the enthalpy of formation of CO2 from diamond from that of graphite.

More Questions from THERMODYNAMICS

Practice All Chapter MCQs →

Consider following reactions
I:

\ \mathrm{C}(\mathrm{s})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})-\mathrm{CO}(\mathrm{g}), \Delta \mathrm{H}_1=\mathrm{a}

II:

\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})-\mathrm{CO}_2(\mathrm{~g}), \Delta \mathrm{H}_2=\mathrm{b}

III:

\mathrm{C}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})-2 \mathrm{CO}(\mathrm{g}) ; \Delta \mathrm{H}_3=\mathrm{c}

Select the correct statement When

0.18 \mathrm{~g}

of glucose was burnt in a bomb calorimeter, the temperature rose by

4 \mathrm{~K}.

The heat capacity of the calorimeter system is

\left(\Delta_{\mathrm{C}} \text{H}^{\circ}=-2.8\times 10^6 \mathrm{~J} / \mathrm{mol} \text{ glucose}\right)

Statement-1:

{CsI}

is less soluble in water compared to

{CsCl}

Satement-2: More amount of heat is liberated during the formation of

{CsCl}

than the formation

{Cs} I

from their elements Consider the following processes :

\begin{array}{cccccccccccccccccccc} {}&{\Delta H(KJ/mol)} \\ {1/2A \to B}&{ + 150} \\ {3B \to 2C + D}&{ - 125} \\ {E + A \to 2D}&{ + 350} \end{array}

For

{B}+{D} \rightarrow {E}+2 {C}, \Delta {H}

will be : The difference b/w

\Delta H

and

\Delta u

at constant volume is equal to The heat evolved in the combustion of methane is given by the following equations

\mathrm{C}{\mathrm{H}_4}\left( \mathrm{g} \right) + 2{\mathrm{O}_2}\left( \mathrm{g} \right) \to \mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right) + {\mathrm{H}_2}\mathrm{O}\left( l \right);\Delta H = - 890.3\mathrm{kJ}

How many grams of methane would be required to produce 44.15 kJ of heat of combustion?