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Question: Write the reaction which has a heat of reaction equal to heat of formation for \(HCl(g)\)?...

Write the reaction which has a heat of reaction equal to heat of formation for HCl(g)HCl(g)?

Explanation

Solution

We know that the amount of heat that should be added or removed during a chemical reaction to keep all the substances at the same temperature is called the heat of reaction. The heat of formation of a chemical reaction is the amount of heat that is absorbed or evolved when a mole of a compound is formed from elements, each element being in its normal physical state.

Complete answer:
One molecule of hydrogen chloride has one atom of HHand one atom of ClCl. We know that hydrogen and chlorine exist as diatomic elements in their standard state. So, the balanced chemical reaction can be written as follows:
H2(g)+Cl2(g)2HCl(g){H_2}(g) + C{l_2}(g) \to 2HCl(g)
For the standard enthalpy change of formation to be equal to the enthalpy change of reaction, we have to divide the chemical reaction by 22. Then the resulting equation is as follows:
12H2(g)+12Cl2(g)HCl(g)\dfrac{1}{2}{H_2}(g) + \dfrac{1}{2}C{l_2}(g) \to HCl(g)
This chemical equation has ΔHrxn=ΔHf\Delta H_{{\text{rxn}}}^ \circ = \Delta H_f^ \circ as it shows the formation of one mole of hydrogen chloride from hydrogen and chlorine.
Therefore, the reaction having a heat of reaction equal to heat of formation for HCl(g)HCl(g) is 12H2(g)+12Cl2(g)HCl(g)\dfrac{1}{2}{H_2}(g) + \dfrac{1}{2}C{l_2}(g) \to HCl(g).

Note:
The standard heat of formation of a compound generally represents the change in enthalpy when one mole of that compound is formed from its constituent elements in their standard state. The change in enthalpy of a chemical reaction describing the formation of one mole of hydrogen chloride from its constituent elements in their standard state is equal to ΔHf\Delta H_f^ \circ .