
The heat of neutralization of LiOH and HCl at ${{25}^{\circ }}C$ is -34.868 KJ /mol. The heat ionization of LiOH will be:
(a)- 44.674 KJ
(b)- 22.232 KJ
(c)- 32.684 KJ
(d)- 96.469 KJ
Answer
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Hint:When a strong acid and strong base combine, there is the formation of water whose heat of formation is equal to -57.1 KJ /mol, which means this is energy is equal to the combining of hydrogen and hydroxyl ions. We know the LiOH is a water base, so the value of ionization will be smaller than this value.
Complete step-by-step answer:The heat of neutralization means when a specific amount of acid is added to the base then there is the formation of salt and water. When the strong acid like HCl and strong base like NaOH, the heat of neutralization is -57.1 KJ /mol. The reaction is given below:
$NaOH+HCl\to NaCl+{{H}_{2}}O\text{ }\Delta \text{H= -57}\text{.1 KJ/mol}$
The formation of water is equal to this value, the reaction can be written as:
${{H}^{+}}+O{{H}^{-}}\to {{H}_{2}}O\text{ }\Delta \text{H= -57}\text{.1 KJ/mol}$
When we reverse this reaction, the value will be:
${{H}_{2}}O\to {{H}^{+}}+O{{H}^{-}}\text{ }\Delta \text{H= 57}\text{.1 KJ/mol}$
Let us say that this is equation 1.
We are given the neutralization of LiOH and HCl is -34.868 KJ /mol. Therefore, the reaction is given below:
${{H}^{+}}+LiOH\to L{{i}^{+}}+{{H}_{2}}O\text{ }\Delta \text{H= -34}\text{.868 KJ/mol}$
Let us say that this equation is 2.
When we add the equation 1 and 2, we can write:
${{H}_{2}}O+LiOH+{{H}^{+}}\to {{H}^{+}}+O{{H}^{-}}+L{{i}^{+}}+{{H}_{2}}O$
Now adding the values of enthalpy of these two reactions, the value will be:
$\Delta H=-34.868+57.1=22.232\text{ KJ /mol}$
The overall reaction will be:
$LiOH\to L{{i}^{+}}+O{{H}^{-}}$
The value of heat of ionization is 22.232 KJ /mol.
Therefore, the correct answer is an option (b).
Note: It must be noted that the highest value of the heat of the water will be -57.1 KJ /mol, if any of the components is changed like strong acid, strong base, weak acid, or weak base, will have a lower value of the heat of the water.
Complete step-by-step answer:The heat of neutralization means when a specific amount of acid is added to the base then there is the formation of salt and water. When the strong acid like HCl and strong base like NaOH, the heat of neutralization is -57.1 KJ /mol. The reaction is given below:
$NaOH+HCl\to NaCl+{{H}_{2}}O\text{ }\Delta \text{H= -57}\text{.1 KJ/mol}$
The formation of water is equal to this value, the reaction can be written as:
${{H}^{+}}+O{{H}^{-}}\to {{H}_{2}}O\text{ }\Delta \text{H= -57}\text{.1 KJ/mol}$
When we reverse this reaction, the value will be:
${{H}_{2}}O\to {{H}^{+}}+O{{H}^{-}}\text{ }\Delta \text{H= 57}\text{.1 KJ/mol}$
Let us say that this is equation 1.
We are given the neutralization of LiOH and HCl is -34.868 KJ /mol. Therefore, the reaction is given below:
${{H}^{+}}+LiOH\to L{{i}^{+}}+{{H}_{2}}O\text{ }\Delta \text{H= -34}\text{.868 KJ/mol}$
Let us say that this equation is 2.
When we add the equation 1 and 2, we can write:
${{H}_{2}}O+LiOH+{{H}^{+}}\to {{H}^{+}}+O{{H}^{-}}+L{{i}^{+}}+{{H}_{2}}O$
Now adding the values of enthalpy of these two reactions, the value will be:
$\Delta H=-34.868+57.1=22.232\text{ KJ /mol}$
The overall reaction will be:
$LiOH\to L{{i}^{+}}+O{{H}^{-}}$
The value of heat of ionization is 22.232 KJ /mol.
Therefore, the correct answer is an option (b).
Note: It must be noted that the highest value of the heat of the water will be -57.1 KJ /mol, if any of the components is changed like strong acid, strong base, weak acid, or weak base, will have a lower value of the heat of the water.
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