The enthalpies of formation of $\mathrm{C}_2 \mathrm{H}_2(\mathrm{~g})$ and $\mathrm{C}_6 \mathrm{H}_6(\mathrm{~g})$ at $298 \mathrm{~K}$ are 230 and $85 \mathrm{~kJ} / \mathrm{mol}$ respectively. The enthalpy change for the reaction is

$865 \mathrm{~kJ} / \mathrm{mol}$

$-865 \mathrm{~kJ} / \mathrm{mol}$

JEE CHEMISTRY MCQ: THERMODYNAMICS

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Application +4 | -1

The enthalpies of formation of

\mathrm{C}_2 \mathrm{H}_2(\mathrm{~g})

and

\mathrm{C}_6 \mathrm{H}_6(\mathrm{~g})

298 \mathrm{~K}

are 230 and

85 \mathrm{~kJ} / \mathrm{mol}

respectively. The enthalpy change for the reaction is

View answer and explanation

Correct option: C

-605 \mathrm{~kJ} / \mathrm{mol}

Option C (-605 kJ/mol) is correct because the enthalpy change for the reaction is calculated by subtracting the sum of the enthalpies of formation of the reactants from the enthalpy of formation of the products.

More Questions from THERMODYNAMICS

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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}

II:

\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})-\mathrm{CO}_2(\mathrm{~g}), \Delta \mathrm{H}_2=\mathrm{b}

III:

\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)

Statement-1:

{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}

For

{B}+{D} \rightarrow {E}+2 {C}, \Delta {H}

will be : The difference b/w

\Delta H

and

\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?