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Hess's law of constant heat summation Short Tricks for JEE

6+ syllabus-aligned questions available

Quick answer

Short tricks for Hess's law of constant heat summation work only with strong fundamentals. Apply the tips below in timed sets and review every explanation.

Use these Hess's law of constant heat summation shortcuts to save time in JEE Chemistry papers — then validate speed with 6+ MCQs on Goodmarks.

Short tricks for speed

  • Hess's law of constant heat summation focus drill

    Solve 15 mixed MCQs for Hess's law of constant heat summation, review every explanation, and note formulas you hesitated on.

  • NCERT line recall

    For Inorganic, exact NCERT phrasing often matches the correct option — highlight one-liners during revision.

  • Balance then compare

    In Physical Chemistry numericals, write the balanced relation first, then substitute values with units.

Syllabus context

Part of Chemical Thermodynamics in JEE Main Chemistry.

Free sample questions

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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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Frequently asked questions

Are short tricks enough for Hess's law of constant heat summation in JEE?

No — tricks complement concepts. Master the theory first, then use shortcuts in timed MCQ practice.

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