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JEE CHEMISTRY MCQ: THERMODYNAMICS

1 Hard

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Analyzing +4 | -1

Statement I Based on the following thermochemical
reactions;

\mathrm{HF}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{H}_2 \mathrm{O}(l)+\mathrm{F}^{-}(a q) ; \Delta H^{\circ}=-16.4 \mathrm{~k~cal}

\mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{H}_2 \mathrm{O}(l), \Delta H^{\circ}=-13.7 \mathrm{~k~cal}

Ionisation of

\mathrm{HF}(\mathrm{aq})

is an exothermic reaction.

Statement II Enthalpy decrease in hydration of

\mathrm{H}^{+}(a q)

and

\mathrm{F}^{-}(a q)

exceeds the ionisation enthalpy of

\mathrm{HF}(a q)

.

View answer and explanation

Correct option: A

Both Statement I and Statement II are correct and
Statement II is the correct explanation of Statement I

Option A (Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I) is correct because the ionization of HF is exothermic due to the high hydration enthalpy of H+ and F- exceeding the ionization enthalpy.

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?