Question
Read the passage given below and answer the following questions from (i) to (v).
In a chemical reaction, reactants are converted into products and is represented by, Reactants → Products The enthalpy change accompanying a reaction is called the reaction enthalpy. The enthalpy change of a chemical reaction, is given by the symbol $\triangle\text{rH}.$
$\triangle\text{rH}$ = (sum of enthalpies of products) – (sum of enthalpies of reactants)
$\sum\limits_\text{t}\text{a}_{\text{t}}\text{H}_\text{products}-\sum\limits_\text{t}\text{b}_\text{t}\text{H}_\text{reactants}$
Here symbol ∑ (sigma) is used for summation and ai and bi are the stoichiometric coefficients of the products and reactants respectively in the balanced chemical equation. For example, for the reaction
$\text{CH}_4(\text{g})+2\text{O}_2(\text{g})\rightarrow\text{CO}_2(\text{g})+2\text{H}_2\text{O}(\text{l})$
$\triangle_\text{r}\text{H}=\sum\limits_\text{t}\text{a}_{\text{t}}\text{H}_\text{products}-\sum\limits_\text{t}\text{b}_\text{t}\text{H}_\text{reactants}$
$=[\text{H}_\text{m}(\text{CO}_2,\text{g})+2\text{H}_\text{m}(\text{H}_2\text{O},\text{l})]-[\text{H}_\text{m}(\text{CH}_4,\text{g})+2\text{H}_\text{m}(\text{O}_2,\text{g})]$
where $H_m$ is the molar enthalpy. Enthalpy change is a very useful quantity. Knowledge of this quantity is required when one needs to plan the heating or cooling required to maintain an industrial chemical reaction at constant temperature. It is also required to calculate temperature dependence of equilibrium constant.
Standard Enthalpy of Reactions Enthalpy of a reaction depends on the conditions under which a reaction is carried out. It is, therefore, necessary that we must specify some standard conditions. The standard enthalpy of reaction is the enthalpy change for a reaction when all the participating substances are in their standard states. The standard state of a substance at a specified temperature is its pure form at 1 bar. For example, the standard state of liquid ethanol at 298K is pure liquid ethanol at 1 bar; standard state of solid iron at 500K is pure iron at 1 bar. Usually data are taken at 298K. Standard conditions are denoted by adding the superscript 0 to the symbol $\triangle\text{H},$ e.g., $\triangle\text{H}^\phi$
Enthalpy Changes during Phase Transformations Phase transformations also involve energy changes. Ice, for example, requires heat for melting. Normally this melting takes place at constant pressure (atmospheric pressure) and during phase change, temperature remains constant (at 273K).
$\text{H}_2\text{O}(\text{s})\rightarrow\text{H}_2\text{O}(\text{l});\triangle_{\text{fus}}\text{H}^\phi=6.00\text{kJ}\ \text{mol}^{-1}$
Here $\triangle\text{vap}\text{H}^\phi$ is enthalpy of fusion in standard state. If water freezes, then process is reversed and equal amount of heat is given off to the surroundings. The enthalpy change that accompanies melting of one mole of a solid substance in standard state is called standard enthalpy of fusion or molar enthalpy of fusion, $\triangle\text{fus}\text{H}0.$Melting of a solid is endothermic, so all enthalpies of fusion are positive. Water requires heat for evaporation. At constant temperature of its boiling point Tb and at constant pressure:
$\text{H}_2\text{O}(\text{l})\rightarrow\text{H}_2\text{O}(\text{g});\triangle_{\text{vap}}\text{H}^\phi=+40.79\text{kJ}\ \text{mol}^{-1}$
$\triangle\text{vap}\text{H}^\phi$ is the standard enthalpy of vaporisation. Amount of heat required to vaporize one mole of a liquid at constant temperature and under standard pressure (1bar) is called its standard enthalpy of vaporization or molar enthalpy of vaporization, $\triangle\text{vap}\text{H}^\phi.$ Sublimation is direct conversion of a solid into its vapour. Solid $CO_2 $or ‘dry ice’ sublimes at 195K with $\triangle\text{sub}\text{H}^\phi=25.2\text{kJ}\text{mol}^{–1};$ naphthalene sublimes slowly and for this $\triangle\text{sub}\text{H}0= 73.0\text{kJ}\text{mol}^{–1}.$ Standard enthalpy of sublimation, $\triangle\text{sub}\text{H}^\phi$ is the change in enthalpy when one mole of a solid substance sublimes at a constant temperature and under standard pressure (1bar). The magnitude of the enthalpy change depends on the strength of the intermolecular interactions in the substance undergoing the phase transfomations. For example, the strong hydrogen bonds between water molecules hold them tightly in liquid phase. For an organic liquid, such as acetone, the intermolecular dipole-dipole interactions are significantly weaker. Thus, it requires less heat to vaporise 1 mol of acetone than it does to vaporize 1mol of water.
Standard Enthalpy of Formation The standard enthalpy change for the formation of one mole of a compound from its elements in their most stable states of aggregation (also known as reference states) is called Standard Molar Enthalpy of Formation. Its symbol is $\triangle\text{f}\text{H}^\phi$ where the subscript ‘ f ’ indicates that one mole of the compound in question has been formed in its standard state from its elements in their most stable states of aggregation. The reference state of an element is its most stable state of aggregation at $25^\circ C$ and 1 bar pressure.
Hess’s Law of Constant Heat Summation We know that enthalpy is a state function, therefore the change in enthalpy is independent of the path between initial state (reactants) and final state (products). In other words, enthalpy change for a reaction is the same whether it occurs in one step or in a series of steps. This may be stated as follows in the form of Hess’s Law. If a reaction takes place in several steps then its standard reaction enthalpy is the sum of the standard enthalpies of the intermediate reactions into which the overall reaction may be divided at the same temperature. Let us understand the importance of this law with the help of an example. Consider the enthalpy change for the reaction
$\text{C}(\text{graphite,s})+\frac{1}{2}\text{O}_2(\text{g})\rightarrow\text{CO}(\text{g});\triangle_\text{r}\text{H}^{\ominus}=?$
Although CO(g) is the major product, some $CO_2 $gas is always produced in this reaction. Therefore, we cannot measure enthalpy change for the above reaction directly. However, if we can find some other reactions involving related species, it is possible to calculate the enthalpy change for the above reaction. Let us consider the following reactions:
$\text{C}(\text{graphite,s})+\text{O}_2(\text{g}) \rightarrow\text{CO}_2(\text{g});\triangle\text{r}\text{H}^{\phi}=–393.5\text{kJ}\text{mol}^{–1}(\text{i})$
$\text{CO}(\text{g})+\frac{1}{2}\text{O}_2(\text{g})\rightarrow\text{CO}_2(\text{g})\triangle_\text{r}\text{H}^{\phi}=-283.0\text{kJ}\text{mol}^{-1}(\text{ii})$
We can combine the above two reactions in such a way so as to obtain the desired reaction. To get one mole of CO(g) on the right, we reverse equation (ii). In this, heat is absorbed instead of being released, so we change sign of $\triangle\text{r}\text{H}^\phi$ value
$\text{CO}_2(\text{g})\rightarrow\text{CO}(\text{g})+\frac{1}{2}\text{O}_2(\text{g});\triangle\text{r}\text{H}^{\phi}=+283.0\text{kJ}\text{mol}^{-1}...(\text{iii})$
Adding equation (i) and (iii), we get the desired equation,
$\text{C}(\text{graphite,s})+\frac{1}{2}\text{O}_2(\text{g})\rightarrow\text{CO}(\text{g});$
for which $\triangle_\text{r}\text{H}^{\phi}=(-393.5+283.0)=-110.5\text{kJ}\text{mol}^{-1}$
In general, if enthalpy of an overall reaction A → B along one route is $\triangle\text{rH}$ and$\triangle\text{rH}_1,\triangle\text{rH}_2,\triangle\text{rH}_3$ representing enthalpies of reactions leading to same product, B along another route, then we have
$\triangle\text{rH}=\triangle\text{rH}_1+\triangle\text{rH}_2+\triangle\text{rH}_3$
It can be represented as:
In a chemical reaction, reactants are converted into products and is represented by, Reactants → Products The enthalpy change accompanying a reaction is called the reaction enthalpy. The enthalpy change of a chemical reaction, is given by the symbol $\triangle\text{rH}.$
$\triangle\text{rH}$ = (sum of enthalpies of products) – (sum of enthalpies of reactants)
$\sum\limits_\text{t}\text{a}_{\text{t}}\text{H}_\text{products}-\sum\limits_\text{t}\text{b}_\text{t}\text{H}_\text{reactants}$
Here symbol ∑ (sigma) is used for summation and ai and bi are the stoichiometric coefficients of the products and reactants respectively in the balanced chemical equation. For example, for the reaction
$\text{CH}_4(\text{g})+2\text{O}_2(\text{g})\rightarrow\text{CO}_2(\text{g})+2\text{H}_2\text{O}(\text{l})$
$\triangle_\text{r}\text{H}=\sum\limits_\text{t}\text{a}_{\text{t}}\text{H}_\text{products}-\sum\limits_\text{t}\text{b}_\text{t}\text{H}_\text{reactants}$
$=[\text{H}_\text{m}(\text{CO}_2,\text{g})+2\text{H}_\text{m}(\text{H}_2\text{O},\text{l})]-[\text{H}_\text{m}(\text{CH}_4,\text{g})+2\text{H}_\text{m}(\text{O}_2,\text{g})]$
where $H_m$ is the molar enthalpy. Enthalpy change is a very useful quantity. Knowledge of this quantity is required when one needs to plan the heating or cooling required to maintain an industrial chemical reaction at constant temperature. It is also required to calculate temperature dependence of equilibrium constant.
Standard Enthalpy of Reactions Enthalpy of a reaction depends on the conditions under which a reaction is carried out. It is, therefore, necessary that we must specify some standard conditions. The standard enthalpy of reaction is the enthalpy change for a reaction when all the participating substances are in their standard states. The standard state of a substance at a specified temperature is its pure form at 1 bar. For example, the standard state of liquid ethanol at 298K is pure liquid ethanol at 1 bar; standard state of solid iron at 500K is pure iron at 1 bar. Usually data are taken at 298K. Standard conditions are denoted by adding the superscript 0 to the symbol $\triangle\text{H},$ e.g., $\triangle\text{H}^\phi$
Enthalpy Changes during Phase Transformations Phase transformations also involve energy changes. Ice, for example, requires heat for melting. Normally this melting takes place at constant pressure (atmospheric pressure) and during phase change, temperature remains constant (at 273K).
$\text{H}_2\text{O}(\text{s})\rightarrow\text{H}_2\text{O}(\text{l});\triangle_{\text{fus}}\text{H}^\phi=6.00\text{kJ}\ \text{mol}^{-1}$
Here $\triangle\text{vap}\text{H}^\phi$ is enthalpy of fusion in standard state. If water freezes, then process is reversed and equal amount of heat is given off to the surroundings. The enthalpy change that accompanies melting of one mole of a solid substance in standard state is called standard enthalpy of fusion or molar enthalpy of fusion, $\triangle\text{fus}\text{H}0.$Melting of a solid is endothermic, so all enthalpies of fusion are positive. Water requires heat for evaporation. At constant temperature of its boiling point Tb and at constant pressure:
$\text{H}_2\text{O}(\text{l})\rightarrow\text{H}_2\text{O}(\text{g});\triangle_{\text{vap}}\text{H}^\phi=+40.79\text{kJ}\ \text{mol}^{-1}$
$\triangle\text{vap}\text{H}^\phi$ is the standard enthalpy of vaporisation. Amount of heat required to vaporize one mole of a liquid at constant temperature and under standard pressure (1bar) is called its standard enthalpy of vaporization or molar enthalpy of vaporization, $\triangle\text{vap}\text{H}^\phi.$ Sublimation is direct conversion of a solid into its vapour. Solid $CO_2 $or ‘dry ice’ sublimes at 195K with $\triangle\text{sub}\text{H}^\phi=25.2\text{kJ}\text{mol}^{–1};$ naphthalene sublimes slowly and for this $\triangle\text{sub}\text{H}0= 73.0\text{kJ}\text{mol}^{–1}.$ Standard enthalpy of sublimation, $\triangle\text{sub}\text{H}^\phi$ is the change in enthalpy when one mole of a solid substance sublimes at a constant temperature and under standard pressure (1bar). The magnitude of the enthalpy change depends on the strength of the intermolecular interactions in the substance undergoing the phase transfomations. For example, the strong hydrogen bonds between water molecules hold them tightly in liquid phase. For an organic liquid, such as acetone, the intermolecular dipole-dipole interactions are significantly weaker. Thus, it requires less heat to vaporise 1 mol of acetone than it does to vaporize 1mol of water.
Standard Enthalpy of Formation The standard enthalpy change for the formation of one mole of a compound from its elements in their most stable states of aggregation (also known as reference states) is called Standard Molar Enthalpy of Formation. Its symbol is $\triangle\text{f}\text{H}^\phi$ where the subscript ‘ f ’ indicates that one mole of the compound in question has been formed in its standard state from its elements in their most stable states of aggregation. The reference state of an element is its most stable state of aggregation at $25^\circ C$ and 1 bar pressure.
Hess’s Law of Constant Heat Summation We know that enthalpy is a state function, therefore the change in enthalpy is independent of the path between initial state (reactants) and final state (products). In other words, enthalpy change for a reaction is the same whether it occurs in one step or in a series of steps. This may be stated as follows in the form of Hess’s Law. If a reaction takes place in several steps then its standard reaction enthalpy is the sum of the standard enthalpies of the intermediate reactions into which the overall reaction may be divided at the same temperature. Let us understand the importance of this law with the help of an example. Consider the enthalpy change for the reaction
$\text{C}(\text{graphite,s})+\frac{1}{2}\text{O}_2(\text{g})\rightarrow\text{CO}(\text{g});\triangle_\text{r}\text{H}^{\ominus}=?$
Although CO(g) is the major product, some $CO_2 $gas is always produced in this reaction. Therefore, we cannot measure enthalpy change for the above reaction directly. However, if we can find some other reactions involving related species, it is possible to calculate the enthalpy change for the above reaction. Let us consider the following reactions:
$\text{C}(\text{graphite,s})+\text{O}_2(\text{g}) \rightarrow\text{CO}_2(\text{g});\triangle\text{r}\text{H}^{\phi}=–393.5\text{kJ}\text{mol}^{–1}(\text{i})$
$\text{CO}(\text{g})+\frac{1}{2}\text{O}_2(\text{g})\rightarrow\text{CO}_2(\text{g})\triangle_\text{r}\text{H}^{\phi}=-283.0\text{kJ}\text{mol}^{-1}(\text{ii})$
We can combine the above two reactions in such a way so as to obtain the desired reaction. To get one mole of CO(g) on the right, we reverse equation (ii). In this, heat is absorbed instead of being released, so we change sign of $\triangle\text{r}\text{H}^\phi$ value
$\text{CO}_2(\text{g})\rightarrow\text{CO}(\text{g})+\frac{1}{2}\text{O}_2(\text{g});\triangle\text{r}\text{H}^{\phi}=+283.0\text{kJ}\text{mol}^{-1}...(\text{iii})$
Adding equation (i) and (iii), we get the desired equation,
$\text{C}(\text{graphite,s})+\frac{1}{2}\text{O}_2(\text{g})\rightarrow\text{CO}(\text{g});$
for which $\triangle_\text{r}\text{H}^{\phi}=(-393.5+283.0)=-110.5\text{kJ}\text{mol}^{-1}$
In general, if enthalpy of an overall reaction A → B along one route is $\triangle\text{rH}$ and$\triangle\text{rH}_1,\triangle\text{rH}_2,\triangle\text{rH}_3$ representing enthalpies of reactions leading to same product, B along another route, then we have
$\triangle\text{rH}=\triangle\text{rH}_1+\triangle\text{rH}_2+\triangle\text{rH}_3$
It can be represented as:
- The enthalpy change of a chemical reaction, is given by the symbol …
- $\triangle\text{rH}$
- $\triangle\text{rG}$
- $\triangle\text{rF}$
- $\triangle\text{rR}$'
- The molar enthalpy is denoted by:
- $H_k$
- $H_m$
- $H_l$
- $H_n$
- …is enthalpy of fusion in standard state.
- $\triangle\text{fus}\text{H}^{\phi}$
- $\triangle_\text{r}\text{H}^{\phi}$
- $\triangle\text{vap}\text{H}^{\phi}$
- $\triangle\text{w}\text{H}^{\phi}$
- Solid $CO_2$or ‘dry ice’ sublimes at..
- $100K$
- $195K$
- $150K$
- $200K$
- … is the standard enthalpy of vaporisation.
- $\triangle\text{fus}\text{H}^{\phi}$
- $\triangle_\text{r}\text{H}^{\phi}$
- $\triangle\text{vap}\text{H}^{\phi}$
- $\triangle\text{w}\text{H}^{\phi}$
