
Enthalpy of phase transition is:
Given that ${{C}_{(graphite)}}\to {{C}_{(g)}};\text{ }\Delta \text{H = +716}\text{.7 KJ}$
${{C}_{(diamond)}}\to {{C}_{(g)}};\text{ }\Delta \text{H = +714}\text{.8 KJ}$. The$\Delta \text{H}$ for the following reaction is ${{C}_{(graphite)}}\to {{C}_{(diamond)}}$ will be:
(a)- 1.9 KJ
(b)- -1.9 KJ
(c)- Zero
(d)- 714.8 KJ
Answer
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Hint: The enthalpy of phase transition of carbon in graphite form to carbon in diamond form can be calculated by the two reactions above. If the reactant is converted into a product and the product is converted into a reactant, the sign of $\Delta \text{H}$will become the opposite.
Complete answer:
There are two reactions given in the question above and these are:
${{C}_{(graphite)}}\to {{C}_{(g)}};\text{ }\Delta \text{H = +716}\text{.7 KJ}$-(i)
${{C}_{(diamond)}}\to {{C}_{(g)}};\text{ }\Delta \text{H = +714}\text{.8 KJ}$-(ii)
In both reactions, the reactant is in solid form and the products are in gas form, so the sublimation reaction is taking place in both the reactions. And both the reactions have the value of enthalpy positive.
We have to find the enthalpy of reaction ${{C}_{(graphite)}}\to {{C}_{(diamond)}}$.
In this reaction, the graphite form is converted into a diamond form. This can be calculated from the given two reactions.
In reaction (i), the graphite is a reactant and the gaseous form is the product. And in ${{C}_{(graphite)}}\to {{C}_{(diamond)}}$, we want graphite as the reactant, so we don’t have to change the reaction.
In reaction (ii), the diamond is a reactant and gaseous form is the product. And in ${{C}_{(graphite)}}\to {{C}_{(diamond)}}$, we want diamond as the product, so we have to reverse the reaction and the sign of the enthalpy for this reaction will become negative.
${{C}_{(g)}}\to {{C}_{(diamond)}};\text{ }\Delta \text{H = -714}\text{.8 KJ}$
So, now adding the reactions we can get the enthalpy of the required reaction, therefore,
$\Delta H=+716.7+(-714.8)$
$\Delta H=1.9\text{ }KJ$
Hence, the correct answer is option (a)- 1.9 KJ.
Note:
Don’t forget to change the sign of the enthalpy when you reverse any reaction. There are three types of enthalpy of phase transition: enthalpy of fusion, enthalpy of vaporization, and enthalpy of sublimation.
Complete answer:
There are two reactions given in the question above and these are:
${{C}_{(graphite)}}\to {{C}_{(g)}};\text{ }\Delta \text{H = +716}\text{.7 KJ}$-(i)
${{C}_{(diamond)}}\to {{C}_{(g)}};\text{ }\Delta \text{H = +714}\text{.8 KJ}$-(ii)
In both reactions, the reactant is in solid form and the products are in gas form, so the sublimation reaction is taking place in both the reactions. And both the reactions have the value of enthalpy positive.
We have to find the enthalpy of reaction ${{C}_{(graphite)}}\to {{C}_{(diamond)}}$.
In this reaction, the graphite form is converted into a diamond form. This can be calculated from the given two reactions.
In reaction (i), the graphite is a reactant and the gaseous form is the product. And in ${{C}_{(graphite)}}\to {{C}_{(diamond)}}$, we want graphite as the reactant, so we don’t have to change the reaction.
In reaction (ii), the diamond is a reactant and gaseous form is the product. And in ${{C}_{(graphite)}}\to {{C}_{(diamond)}}$, we want diamond as the product, so we have to reverse the reaction and the sign of the enthalpy for this reaction will become negative.
${{C}_{(g)}}\to {{C}_{(diamond)}};\text{ }\Delta \text{H = -714}\text{.8 KJ}$
So, now adding the reactions we can get the enthalpy of the required reaction, therefore,
$\Delta H=+716.7+(-714.8)$
$\Delta H=1.9\text{ }KJ$
Hence, the correct answer is option (a)- 1.9 KJ.
Note:
Don’t forget to change the sign of the enthalpy when you reverse any reaction. There are three types of enthalpy of phase transition: enthalpy of fusion, enthalpy of vaporization, and enthalpy of sublimation.
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