Question

A ${\text{38}}{\text{.55kJ}}$ of heat absorbed when ${\text{6g}}$ of ${{\text{O}}_{\text{2}}}$ react with ClF according to the reaction-
$2ClF(g) + {O_2} \to C{l_2}O(g) + $$O{F_2}(g)$
What is the standard enthalpy of the reaction? ($\Delta {{\text{H}}_{\text{o}}}{\text{ = + 205}}{\text{.6kJ}}$)

Answer 1

Hint:In the above question, we have to first check if the equation given is balanced or not. Then only we can find out the standard enthalpy of a reaction. Since we are provided with ${{\text{O}}_{\text{2}}}$ weight, we will consider how much heat is absorbed for 1 mole of oxygen to find out the standard enthalpy of the reaction.

Complete step by step answer:
Standard enthalpy of reaction is defined as the enthalpy change when the product and react are in their standard state. It is often referred to as per mole reaction in a balanced equation.
So we are given the equation:
$2ClF(g) + {O_2} \to C{l_2}O(g) + $$O{F_2}(g)$
Step 1:
 First of all, we have to check whether it is a balanced equation or not. Since, we have 2Cl, 2F and 2O on both sides of the reaction. So, now we can proceed further.
Step 2:
Since we are provided with ${{\text{O}}_{\text{2}}}$weight, so we have to find the number of mole of ${{\text{O}}_{\text{2}}}$reacted to find out the standard enthalpy .
No. of moles of ${{\text{O}}_{\text{2}}}$(n)=$\dfrac{{\text{m}}}{{\text{M}}}$
where m= given mass of ${{\text{O}}_{\text{2}}}$
 M= molar mass of ${{\text{O}}_{\text{2}}}$
m=6 (given)
M= 2$ \times $atomic mass of ${{\text{O}}_{\text{2}}}$=2$ \times $16 =32
So, ${\text{n = }}\dfrac{{\text{6}}}{{{\text{32}}}}{\text{ = 0}}{\text{.1875 moles}}$
So, when ${\text{0}}{\text{.1875 moles}}$ of ${{\text{O}}_{\text{2}}}$is reacted then ${\text{38}}{\text{.55 kJ}}$ of heat is absorbed.
But according to the balanced equation:
$2ClF(g) + {O_2} \to C{l_2}O(g) + $$O{F_2}(g)$
We are interested to find out how much energy is absorbed when $1{\text{ mole}}$ of O$_2$is reacted so ${\text{0}}{\text{.1875 moles}}$ of ${{\text{O}}_{\text{2}}}$${\text{ = 38}}{\text{.55 kJ}}$ of heat absorbed
$ \Rightarrow 1{\text{ mole of }}{{\text{O}}_2} = \dfrac{{{\text{38}}{\text{.55}}}}{{{\text{0}}{\text{.1875}}}}{\text{kJ}} = 205.6{\text{ kJ}}$ of heat absorbed

$\therefore $Standard heat of enthalpy is ${\text{205}}{\text{.6 kJ}}$

Note:
Suppose we have a balanced equation as given below:
A$ \times $(Reactant 1) + B$ \times $(Reactant 2)$ \to $ C$ \times $(Product 1)+ D$ \times $(Product 2)
Standard enthalpy of reaction is the same as the enthalpy required for the reaction of A moles of Reactant 1 or B moles of Reactant 2 or the formation of C moles of Product 1 or the D moles of Product 2. Since the number of moles in the reactant and the product are interrelated in a balanced equation so knowing about one of them is equivalent to knowing all of them. Hence this is also called mole reaction.