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Enthalpy Change - Practice Questions & MCQ
Edited By admin | Updated on Sep 18, 2023 18:35 AM | #JEE Main
Quick Facts
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38 Questions around this concept.
Solve by difficulty
Given C(graphite)+O2(g) → CO2(g) ;
rH0=−393.5 kJ mol−1
H2(g)+ O2(g) → H2O(l) ;
rH0=−285.8 kJ mol−1
CO2(g)+2H2O(l) → CH4(g)+2O2(g) ;
rH0=+890.3 kJ mol−1
Based on the above thermochemical equations, the value of rH0 at 298 K for the reaction
C(graphite)+2H2(g) → CH4(g) will be :
The standard enthalpy change for a reaction is
. Calculate the heat (q) released when 2 moles of the reaction take place at constant pressure.
An ideal gas undergoes a reversible isobaric expansion at a constant pressure of . The gas absorbs
of heat and its temperature increases by
. Calculate the change in heat enthalpy of the gas. Given the heat capacity at constant pressure
for the gas is
An ideal gas undergoes a reversible isobaric expansion at a constant pressure of . The gas absorbs
of heat and its temperature increases by
. Calculate the change in heat enthalpy of the gas. Given the heat capacity at constant pressure
for the gas is
An ideal gas undergoes a reversible isobaric expansion at a constant pressure of The gas absorbs
of heat and its temperature increases by
. Calculate the change in heat enthalpy of the gas. Given the heat capacity at constant pressure
for the gas is
A chemical reaction is carried out in a bomb calorimeter. The temperature of the calorimeter increases from upon the reaction. The heat capacity of the calorimeter (including the water) is
. Calculate the heat released
by the reaction.
When 9 in ice melts at 00 C, x KJ heat is required, then enthalpy of fusion of ice will be
Using the following reactions and their respective enthalpy changes, calculate the enthalpy change for the reaction:
$
2 \mathrm{C}(\text { graphite })+\mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{CO}(\mathrm{~g})
$
Reactions:
1. C (graphite) $+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})$ with $\Delta \mathrm{H}^{\circ}=-393.5 \mathrm{~kJ} / \mathrm{mol}$
2. $\mathrm{CO}_2(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})$ with $\Delta \mathrm{H}^{\circ}=-283.0 \mathrm{~kJ} / \mathrm{mol}$
Concepts Covered - 1
Standard Enthalpy And Enthalpy Of Formation
Heat of Formation
The amount of heat evolved or absorbed or change in enthalpy when 1 mole of a substance is obtained from its constituents or free elements.
$\Delta \mathrm{H}^{\circ}=\sum \mathrm{H}^{\circ} \mathrm{p}-\sum \mathrm{H}^{\circ} \mathrm{r}$
Once the value of Ho at 25o C for any species has been assigned the value of Ho at other temperatures Can be found out by using Kirchoff's equation as follows.
$\int_{298}^{\mathrm{T}} \mathrm{dH}^{\circ}=\int_{298}^{\mathrm{T}} \mathrm{CpdT}$
$\mathrm{H}_{\mathrm{T}}^{\circ}-\mathrm{H}_{298}^{\circ}=\int_{298}^{\mathrm{T}} \mathrm{CpdT}$
1. Standard heat of formation of a free element is taken as zero.
For example, In carbon—graphite form is taken as standard state and in sulphur, the monoclinic form is the standard state.
2. The heat of formation may be +ve or -ve.
If $\Delta \mathrm{H}$ is -ve compound is exothermic.
If $\Delta \mathrm{H}$ is + ve compound is endothermic.
3. The stability of the exothermic compound is more than that of the endothermic compound hence, the greater the liberated energy greater is the stability of the compound.
Example, HF > HCI > HBr > HI
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