The heat of combustion of yellow phosphorus is $ - 9.91KJ$ and heat of combustion of red phosphorus is $ - 8.78KJ$ . Which of the following represents the heat of transition of yellow phosphorus to red phosphorus?
A. $ - 18.69KJ$
B. $ + 1.13KJ$
C. $ + 18.69KJ$
D. $ - 1.13KJ$
Answer
615.3k+ views
Hint:We know that enthalpy of a system is the sum of internal energy and product of pressure and volume of that system. Hess’ law used in thermodynamics states that the change of enthalpy in a chemical reaction is a state function and is therefore independent of the path taken between initial and final states. If the enthalpy change for each equation is known, the resultant enthalpy is equal to the sum of enthalpies of each individual reaction.
Formula Used:
We will use the formula,
${\Delta _{net}}H = \sum {\Delta {H_r}} $
${\Delta _{net}}H = $ Net enthalpy of the reaction
$\sum {} {\Delta _r}H = $ Sum of enthalpies of the reactions.
Complete answer:
Phosphorus is a group $15$ element and is represented by the symbol $P$ . It is a $p - $block element and exists in many allotropic forms. Each allotropic form has a different chemical arrangement and different chemical properties. The two most common allotropic forms of phosphorus are white phosphorus and red phosphorus. Red phosphorus is formed by exposing white phosphorus to sunlight. Yellow phosphorus is formed from a small amount of red phosphorus and is highly toxic.
The heat of combustion is the amount of heat released when $1$ mole of a compound is completely burnt. Since it is an exothermic reaction $\Delta H$ is negative. According to the given question
The heat of combustion of yellow phosphorus is $ - 9.91KJ$
$${P_4}\left( y \right) + 5{O_2} \to 2{P_2}{O_5}$$
The heat of combustion of red phosphorus is $ - 8.78KJ$
$${P_4}\left( r \right) + 5{O_2} \to 2{P_2}{O_5}$$
Subtracting both the equations and applying Hess law, we get
${P_4}(y) \to {P_4}(r)$
$\Rightarrow - 9.91 - ( - 8.78) KJ = - 1.13KJ $
Thus heat of transition of yellow phosphorus to red phosphorus is $ - 1.13KJ$ . Therefore, the correct option is D.
Note:
It is important to note that the transition takes place from yellow to red phosphorus. The enthalpy change should be calculated accordingly. Hess’ law is a very useful equation used to calculate the change in enthalpy from a large range of data. Born-Haber cycle is an application of Hess’ law. It is used to calculate the lattice energy of ionic compounds.
Formula Used:
We will use the formula,
${\Delta _{net}}H = \sum {\Delta {H_r}} $
${\Delta _{net}}H = $ Net enthalpy of the reaction
$\sum {} {\Delta _r}H = $ Sum of enthalpies of the reactions.
Complete answer:
Phosphorus is a group $15$ element and is represented by the symbol $P$ . It is a $p - $block element and exists in many allotropic forms. Each allotropic form has a different chemical arrangement and different chemical properties. The two most common allotropic forms of phosphorus are white phosphorus and red phosphorus. Red phosphorus is formed by exposing white phosphorus to sunlight. Yellow phosphorus is formed from a small amount of red phosphorus and is highly toxic.
The heat of combustion is the amount of heat released when $1$ mole of a compound is completely burnt. Since it is an exothermic reaction $\Delta H$ is negative. According to the given question
The heat of combustion of yellow phosphorus is $ - 9.91KJ$
$${P_4}\left( y \right) + 5{O_2} \to 2{P_2}{O_5}$$
The heat of combustion of red phosphorus is $ - 8.78KJ$
$${P_4}\left( r \right) + 5{O_2} \to 2{P_2}{O_5}$$
Subtracting both the equations and applying Hess law, we get
${P_4}(y) \to {P_4}(r)$
$\Rightarrow - 9.91 - ( - 8.78) KJ = - 1.13KJ $
Thus heat of transition of yellow phosphorus to red phosphorus is $ - 1.13KJ$ . Therefore, the correct option is D.
Note:
It is important to note that the transition takes place from yellow to red phosphorus. The enthalpy change should be calculated accordingly. Hess’ law is a very useful equation used to calculate the change in enthalpy from a large range of data. Born-Haber cycle is an application of Hess’ law. It is used to calculate the lattice energy of ionic compounds.
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