
Hess’s law is applicable for the determination of heat of:
A. transition
B. formation
C. reaction
D. all of these
Answer
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Hint: As we know that energy nor be created or destroyed but it can be transferred from one form to another, this is the law of conservation of energy, which is also called the first law of thermodynamics. Hess’s law is also found to be expressed in the first law of thermodynamics.
Complete step by step solution:
- Hess’s law states that during a reaction the total enthalpy change is the same whether the reaction takes place in a single step or in several steps.
- For example, when carbon in a solid-state is heated with oxygen gives carbon-dioxide. Here, we can see that the change in enthalpy is -94Kcal.
\[C(s)+{{O}_{2}}(g)\to C{{O}_{2}}(g)\text{ }\Delta \text{H= -94Kcal}\]
- If this same reaction takes place in many steps, then in the first step carbon is converted into carbon monoxide, then further it is oxidised to carbon dioxide. We can see here that the change in enthalpy that takes place in the first step is -26.4K and in the next step is -67.6K.
\[\begin{align}
& C(s)+\dfrac{1}{2}{{O}_{2}}(g)\to C{{O}_{2}}(g)\text{ }\Delta {{\text{H}}_{1}}\text{= -26}\text{.4Kcal} \\
& C(g)+\dfrac{1}{2}{{O}_{2}}(g)\to C{{O}_{2}}(g)\text{ }\Delta {{\text{H}}_{2}}\text{= -67}\text{.6Kcal} \\
& \text{ }\Delta \text{H=}\Delta {{\text{H}}_{1}}+\Delta {{\text{H}}_{2}} \\
& -94=-26.4-67.6 \\
& -94=-94 \\
\end{align}\]
- Now, as we know that enthalpy change is found to be the same whether the reaction takes place in one or two steps.
Hence, we can conclude that the correct option is (d), that is Hess’s law is applicable for the determination of heat of transition, formation, and reaction.
Note: We can calculate the heat of formation of intermediate compounds if the change in enthalpy for the direct step and final step are known. This law is useful to calculate heat of reaction, lattice energy of crystal, heat of transition.
Complete step by step solution:
- Hess’s law states that during a reaction the total enthalpy change is the same whether the reaction takes place in a single step or in several steps.
- For example, when carbon in a solid-state is heated with oxygen gives carbon-dioxide. Here, we can see that the change in enthalpy is -94Kcal.
\[C(s)+{{O}_{2}}(g)\to C{{O}_{2}}(g)\text{ }\Delta \text{H= -94Kcal}\]
- If this same reaction takes place in many steps, then in the first step carbon is converted into carbon monoxide, then further it is oxidised to carbon dioxide. We can see here that the change in enthalpy that takes place in the first step is -26.4K and in the next step is -67.6K.
\[\begin{align}
& C(s)+\dfrac{1}{2}{{O}_{2}}(g)\to C{{O}_{2}}(g)\text{ }\Delta {{\text{H}}_{1}}\text{= -26}\text{.4Kcal} \\
& C(g)+\dfrac{1}{2}{{O}_{2}}(g)\to C{{O}_{2}}(g)\text{ }\Delta {{\text{H}}_{2}}\text{= -67}\text{.6Kcal} \\
& \text{ }\Delta \text{H=}\Delta {{\text{H}}_{1}}+\Delta {{\text{H}}_{2}} \\
& -94=-26.4-67.6 \\
& -94=-94 \\
\end{align}\]
- Now, as we know that enthalpy change is found to be the same whether the reaction takes place in one or two steps.
Hence, we can conclude that the correct option is (d), that is Hess’s law is applicable for the determination of heat of transition, formation, and reaction.
Note: We can calculate the heat of formation of intermediate compounds if the change in enthalpy for the direct step and final step are known. This law is useful to calculate heat of reaction, lattice energy of crystal, heat of transition.
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