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Hard Hess's law of constant heat summation MCQs for JEE

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Part of Chemical Thermodynamics in JEE Main Chemistry.

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Q1ChemistryUnit 4: Chemical Thermodynamics
Heat of combustion of CH4,C2H6\boldsymbol{C H}_{4}, \boldsymbol{C}_{2} \boldsymbol{H}_{6} and C3H8C_{3} H_{8} are respectively -210,-368.4 and −526.8Kcal-526.8 \mathrm{Kcal} mol 1_{1} Hence, heat of combustion of C8H16C_{8} H_{16} is approximately:
Q2ChemistryUnit 4: Chemical Thermodynamics
The statement "The change of enthalpy of a chemical reaction is same whether the reaction takes place in one or several steps" is:
Q3ChemistryUnit 4: Chemical Thermodynamics
If the heats of formation of C2H2C_{2} H_{2} and C6H6C_{6} H_{6} are 230KJmol−1230 \mathrm{KJ} \mathrm{mol}^{-1} and 85KJmol−185 \mathrm{KJ} \mathrm{mol}^{-1} respectively, the ΔH\Delta \mathrm{H} value for the trimerisation of C2H2C_{2} \mathrm{H}_{2} is:
Q4ChemistryUnit 4: Chemical Thermodynamics
(i) CaO(s)+H2O(l)=Ca(OH)2(s)\boldsymbol{C a O}(\boldsymbol{s})+\boldsymbol{H}_{2} \boldsymbol{O}(l)=\boldsymbol{C a}(\boldsymbol{O H})_{2}(\boldsymbol{s}) ΔH180∘C=−15.26kcal\Delta H_{180^{\circ} C}=-15.26 k c a l (ii) H2O(l)=H2(g)+12O2(g)\boldsymbol{H}_{2} \boldsymbol{O}(\boldsymbol{l})=\boldsymbol{H}_{2}(\boldsymbol{g})+\frac{1}{2} \boldsymbol{O}_{2}(\boldsymbol{g}) ΔH180∘C=68.37kcal\boldsymbol{\Delta} \boldsymbol{H}_{180^{\circ} C}=\mathbf{6 8 . 3 7} \boldsymbol{k c a l} (iii) Ca(s)+12O2(g)=CaO(s)C a(s)+\frac{1}{2} O_{2}(g)=C a O(s) ΔH180∘C=−151.80kcal\boldsymbol{\Delta} \boldsymbol{H}_{180^{\circ} C}=-151.80 k c a l From the following data, the heat of formation of Ca(OH)2(s)\boldsymbol{C a}(\boldsymbol{O H})_{2}(\boldsymbol{s}) at 18∘C\mathbf{1 8}^{\circ} \boldsymbol{C} is:
Q5ChemistryUnit 4: Chemical Thermodynamics
Given the following: C(s)+O2(g)→CO2(g);ΔH=\boldsymbol{C}(s)+\boldsymbol{O}_{2}(\boldsymbol{g}) \rightarrow \boldsymbol{C} \boldsymbol{O}_{2}(\boldsymbol{g}) ; \boldsymbol{\Delta} \boldsymbol{H}= −394kJ/mol-394 k J / m o l 2H2(g)+O2(g)→2H2O(l);ΔH=2 H_{2}(g)+O_{2}(g) \rightarrow 2 H_{2} O(l) ; \Delta H= −568kJ/mol⁡C2H5OH(l)+-\mathbf{5 6 8} k J / \operatorname{mol} C_{2} H_{5} O H(l)+ 3O2(g)→2CO2(g)+\mathbf{3} O_{2}(\boldsymbol{g}) \rightarrow \mathbf{2} \boldsymbol{C} \boldsymbol{O}_{2}(\boldsymbol{g})+ 3H2O(l);ΔH=−1058kJ/mol\mathbf{3} \boldsymbol{H}_{2} \boldsymbol{O}(\boldsymbol{l}) ; \boldsymbol{\Delta} \boldsymbol{H}=-\mathbf{1 0 5 8} \boldsymbol{k} \boldsymbol{J} / \boldsymbol{m o l} Using the given data, the heat of formation of ethanol is:
Q6ChemistryUnit 4: Chemical Thermodynamics
S+32O2→SO3+2x\boldsymbol{S}+\frac{\mathbf{3}}{\mathbf{2}} \boldsymbol{O}_{2} \rightarrow \boldsymbol{S} \boldsymbol{O}_{3}+\mathbf{2} \boldsymbol{x} kcal SO2+12O2→SO3+y\boldsymbol{S O}_{2}+\frac{1}{2} \boldsymbol{O}_{2} \rightarrow \boldsymbol{S} \boldsymbol{O}_{3}+\boldsymbol{y} kcal The heat of formation of SO2S O_{2} is:

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