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JEE CHEMISTRY MCQ: THERMODYNAMICS

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Analyzing +4 | -1
Given below are two statements one is labelled as Assertion (A) and the other is labelled as Reason (R)

Assertion (A): Heat of neutralisation of HFHF (aq.) with NaOH{NaOH} (aq.) is greater than 13.7 Kcal13.7 \mathrm{~Kcal}.

Reason (R): Generally some heat is consumed in the ionisation of weak acid.
On the basis of above statements choose the most appropriate option from the following options:
View answer and explanation
Correct option: A
Both (A) and (R) are correct but (R) is not the correct explanation of (A)
Option A is correct because both the assertion and the reason are correct, but the reason does not explain the assertion. The higher heat of neutralization for HF is due to hydration energy, not just ionization.

More Questions from THERMODYNAMICS

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Consider following reactions
I:  C(s)+12O2( g)−CO(g),ΔH1=a\ \mathrm{C}(\mathrm{s})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})-\mathrm{CO}(\mathrm{g}), \Delta \mathrm{H}_1=\mathrm{a}

II: CO(g)+12O2( g)−CO2( g),ΔH2=b\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})-\mathrm{CO}_2(\mathrm{~g}), \Delta \mathrm{H}_2=\mathrm{b}

III: C(s)+CO2( g)−2CO(g);ΔH3=c\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 g0.18 \mathrm{~g} of glucose was burnt in a bomb calorimeter, the temperature rose by 4 K.4 \mathrm{~K}. The heat capacity of the calorimeter system is (ΔCH∘=−2.8×106 J/mol glucose)\left(\Delta_{\mathrm{C}} \text{H}^{\circ}=-2.8\times 10^6 \mathrm{~J} / \mathrm{mol} \text{ glucose}\right) Statement-1: CsI{CsI} is less soluble in water compared to CsCl{CsCl}
Satement-2: More amount of heat is liberated during the formation of CsCl{CsCl} than the formation CsI{Cs} I from their elements
Consider the following processes :

ΔH(KJ/mol)1/2A→B+1503B→2C+D−125E+A→2D+350\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→E+2C,ΔH{B}+{D} \rightarrow {E}+2 {C}, \Delta {H} will be :
The difference b/w ΔH\Delta H and Δu\Delta u at constant volume is equal to The heat evolved in the combustion of methane is given by the following equations CH4(g)+2O2(g)→CO2(g)+H2O(l);ΔH=−890.3kJ\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?
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