question 23 of 27\nduring a reaction, the enthalpy of formation of an intermediate is 90.3 kj/mol. during…

question 23 of 27\nduring a reaction, the enthalpy of formation of an intermediate is 90.3 kj/mol. during the reaction, 2 moles of the intermediate are formed as a reactant. what is the enthalpy value for this step of the reaction?\na. -90.3 kj\nb. 180.6 kj\nc. 90.3 kj\nd. -180.6 kj

question 23 of 27\nduring a reaction, the enthalpy of formation of an intermediate is 90.3 kj/mol. during the reaction, 2 moles of the intermediate are formed as a reactant. what is the enthalpy value for this step of the reaction?\na. -90.3 kj\nb. 180.6 kj\nc. 90.3 kj\nd. -180.6 kj

Answer

Explanation:

Step1: Recall enthalpy - mole relationship

The enthalpy change ($\Delta H$) for the formation of a substance is related to the number of moles ($n$) and the enthalpy of formation ($\Delta H_f$) by the formula $\Delta H=n\times\Delta H_f$.

Step2: Determine the values of $n$ and $\Delta H_f$

Given that $\Delta H_f = 90.3$ kJ/mol and $n = 2$ mol. Also, since the intermediate is formed as a reactant, the sign of the enthalpy change will be negative (opposite to the enthalpy of formation when it is a product - formation process).

Step3: Calculate the enthalpy change

$\Delta H=-n\times\Delta H_f=- 2\times90.3$ kJ/mol$=-180.6$ kJ.

Answer:

D. -180.6 kJ