(A) For an exothermic reaction,the enthalpy change is negative $(\Delta H < 0)$.$\Delta H = H_P - H_R$,where $H_P$ is the enthalpy of products and $H_

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(A) For an exothermic reaction,the enthalpy change is negative $(\Delta H $\Delta H = H_P - H_R$,where $H_P$ is the enthalpy of products and $H_R$ is the enthalpy of reactants.Since $\Delta H = -74.8 \ kJ \ mol^{-1}$,it implies $H_R > H_P$.Therefore,the energy level of the reactants $(R)$ must be higher than the energy level of the products $(P)$,and the difference between them is $74.8 \ kJ \ mol^{-1}$.Diagram $A$ correctly shows $H_R > H_P$ with the energy difference of $74.8 \ kJ \ mol^{-1}$ representing the enthalpy change.

Class 11 Chemistry Thermodynamics Mix Examples-Thermodynamics and Thermochemistry

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$C_{(s)} + 2 H_{2(g)} \rightarrow CH_{4(g)}$; $\Delta H = -74.8 \ kJ \ mol^{-1}$
Which of the following diagrams gives an accurate representation of the above reaction?
[$R \rightarrow$ reactants; $P \rightarrow$ products]

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$C_{(graphite)} + O_{2(g)} \to CO_{2(g)}; \Delta H = - 94.05 \ k \ cal \ mol^{-1}$
$C_{(diamond)} + O_{2(g)} \to CO_{2(g)}; \Delta H = - 94.50 \ k \ cal \ mol^{-1}$
Therefore:

Medium

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For the reaction
$A_{(\ell)} \rightarrow 2 B_{(g)}$
$\Delta U = 2.1 \; kcal, \Delta S = 20 \; cal \; K^{-1} \; mol^{-1}$ at $300 \; K$
Hence $\Delta G$ in $kcal \; mol^{-1}$ is

DifficultJEE MAIN 2020

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Consider the following statements:
Statement-$I$: During isothermal expansion of an ideal gas,its enthalpy decreases.
Statement-$II$: When $2.0 \ L$ of an ideal gas expands isothermally into vacuum,$\Delta U = 0$.

MediumAP EAMCET 2025

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For the following combustion reaction occurring in an automobile: $2C_{8}H_{18(g)} + 25O_{2(g)} \to 16CO_{2(g)} + 18H_{2}O_{(g)}$,what are the signs of $\Delta H$,$\Delta S$,and $\Delta G$?

MediumAIPMT 1994

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The heats of combustion of $C_2H_4, C_2H_6$ and $H_2$ are $-1409.5 \, kJ, -1558.3 \, kJ$ and $-285.6 \, kJ$ respectively. The heat of hydrogenation of ethene is ...... $kJ$.

Medium

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$C_{(s)} + 2 H_{2(g)} \rightarrow CH_{4(g)}$; $\Delta H = -74.8 \ kJ \ mol^{-1}$Which of the following diagrams gives an accurate representation of the above reaction?[$R \rightarrow$ reactants; $P \rightarrow$ products]

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$C_{(graphite)} + O_{2(g)} \to CO_{2(g)}; \Delta H = - 94.05 \ k \ cal \ mol^{-1}$$C_{(diamond)} + O_{2(g)} \to CO_{2(g)}; \Delta H = - 94.50 \ k \ cal \ mol^{-1}$Therefore:

For the reaction$A_{(\ell)} \rightarrow 2 B_{(g)}$$\Delta U = 2.1 \; kcal, \Delta S = 20 \; cal \; K^{-1} \; mol^{-1}$ at $300 \; K$Hence $\Delta G$ in $kcal \; mol^{-1}$ is

Consider the following statements:Statement-$I$: During isothermal expansion of an ideal gas,its enthalpy decreases.Statement-$II$: When $2.0 \ L$ of an ideal gas expands isothermally into vacuum,$\Delta U = 0$.

For the following combustion reaction occurring in an automobile: $2C_{8}H_{18(g)} + 25O_{2(g)} \to 16CO_{2(g)} + 18H_{2}O_{(g)}$,what are the signs of $\Delta H$,$\Delta S$,and $\Delta G$?

The heats of combustion of $C_2H_4, C_2H_6$ and $H_2$ are $-1409.5 \, kJ, -1558.3 \, kJ$ and $-285.6 \, kJ$ respectively. The heat of hydrogenation of ethene is ...... $kJ$.

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$C_{(s)} + 2 H_{2(g)} \rightarrow CH_{4(g)}$; $\Delta H = -74.8 \ kJ \ mol^{-1}$
Which of the following diagrams gives an accurate representation of the above reaction?
[$R \rightarrow$ reactants; $P \rightarrow$ products]

$C_{(graphite)} + O_{2(g)} \to CO_{2(g)}; \Delta H = - 94.05 \ k \ cal \ mol^{-1}$
$C_{(diamond)} + O_{2(g)} \to CO_{2(g)}; \Delta H = - 94.50 \ k \ cal \ mol^{-1}$
Therefore:

For the reaction
$A_{(\ell)} \rightarrow 2 B_{(g)}$
$\Delta U = 2.1 \; kcal, \Delta S = 20 \; cal \; K^{-1} \; mol^{-1}$ at $300 \; K$
Hence $\Delta G$ in $kcal \; mol^{-1}$ is

Consider the following statements:
Statement-$I$: During isothermal expansion of an ideal gas,its enthalpy decreases.
Statement-$II$: When $2.0 \ L$ of an ideal gas expands isothermally into vacuum,$\Delta U = 0$.