Chemistry · JEE

Hess's law of constant heat summation Mock Test for JEE

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Syllabus context

Part of Chemical Thermodynamics in JEE Main Chemistry.

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Q1ChemistryUnit 4: Chemical Thermodynamics

Heat of combustion of

\boldsymbol{C H}_{4}, \boldsymbol{C}_{2} \boldsymbol{H}_{6}

and

C_{3} H_{8}

are respectively -210,-368.4 and

-526.8 \mathrm{Kcal}

mol

_{1}

Hence, heat of combustion of

C_{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

C_{2} H_{2}

and

C_{6} H_{6}

are

230 \mathrm{KJ} \mathrm{mol}^{-1}

and

85 \mathrm{KJ} \mathrm{mol}^{-1}

respectively, the

\Delta \mathrm{H}

value for the trimerisation of

C_{2} \mathrm{H}_{2}

is:

Q4ChemistryUnit 4: Chemical Thermodynamics

(i)

\boldsymbol{C a O}(\boldsymbol{s})+\boldsymbol{H}_{2} \boldsymbol{O}(l)=\boldsymbol{C a}(\boldsymbol{O H})_{2}(\boldsymbol{s})

\Delta H_{180^{\circ} C}=-15.26 k c a l

(ii)

\boldsymbol{H}_{2} \boldsymbol{O}(\boldsymbol{l})=\boldsymbol{H}_{2}(\boldsymbol{g})+\frac{1}{2} \boldsymbol{O}_{2}(\boldsymbol{g})

\boldsymbol{\Delta} \boldsymbol{H}_{180^{\circ} C}=\mathbf{6 8 . 3 7} \boldsymbol{k c a l}

(iii)

C a(s)+\frac{1}{2} O_{2}(g)=C a O(s)

\boldsymbol{\Delta} \boldsymbol{H}_{180^{\circ} C}=-151.80 k c a l

From the following data, the heat of formation of

\boldsymbol{C a}(\boldsymbol{O H})_{2}(\boldsymbol{s})

\mathbf{1 8}^{\circ} \boldsymbol{C}

is:

Q5ChemistryUnit 4: Chemical Thermodynamics

Given the following:

\boldsymbol{C}(s)+\boldsymbol{O}_{2}(\boldsymbol{g}) \rightarrow \boldsymbol{C} \boldsymbol{O}_{2}(\boldsymbol{g}) ; \boldsymbol{\Delta} \boldsymbol{H}=

-394 k J / m o l

2 H_{2}(g)+O_{2}(g) \rightarrow 2 H_{2} O(l) ; \Delta H=

-\mathbf{5 6 8} k J / \operatorname{mol} C_{2} H_{5} O H(l)+

\mathbf{3} O_{2}(\boldsymbol{g}) \rightarrow \mathbf{2} \boldsymbol{C} \boldsymbol{O}_{2}(\boldsymbol{g})+

\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

\boldsymbol{S}+\frac{\mathbf{3}}{\mathbf{2}} \boldsymbol{O}_{2} \rightarrow \boldsymbol{S} \boldsymbol{O}_{3}+\mathbf{2} \boldsymbol{x}

kcal

\boldsymbol{S O}_{2}+\frac{1}{2} \boldsymbol{O}_{2} \rightarrow \boldsymbol{S} \boldsymbol{O}_{3}+\boldsymbol{y}

kcal The heat of formation of

S O_{2}

is:

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