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JEE CHEMISTRY MCQ: THERMODYNAMICS

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Application +4 | -1

The standard state Gibbs free energies of formation of

C

(graphite) and C (diamond) at

T=298 {~K}

are

\begin{aligned} & \Delta_f G^{\circ}[{C} \text { (graphite) }]=0 {~kJ} {~mol}^{-1} \\ & \Delta_f G^{\circ}[{C} \text { (diamond) }]=2.9 {~kJ} {~mol}^{-1} \end{aligned}

The standard state means that the pressure should be 1 bar, and substance should be pure at a given temperature. The conversion of graphite [C(graphite)] to diamond

\left[C\right.

(diamond)] reduces its volume by

2 \times 10^{-6} {~m}^3 {~mol}^{-1}

. If

{C}

(graphite) is converted to

{C}

(diamond) isothermally at

{T}

=298 {~K}

, the pressure at which

{C}

(graphite) is in equilibrium with

{C}

(diamond), is :

[\text{Useful information:}

1 {~J}=1 {~kg} {~m}^2 {~s}^{-2}

;

\left.1 {~Pa}=1 {~kg} {~m}^{-1} {~s}^{-2} ; 1 {bar}=10^5 {~Pa}\right]

View answer and explanation

Correct option: A

14501 bar

Option A (14501 bar) is correct because it is the pressure calculated using the Gibbs free energy change and volume change for the conversion of graphite to diamond.

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}

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

\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)

{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}

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

will be : The difference b/w

\Delta H

\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?