Question

question 10 of 30
during 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?

a. -90.3 kj

b. 180.6 kj

c. -180.6 kj

d. 90.3 kj

Explanation:

Step1: Recall enthalpy of formation concept

The enthalpy of formation ($\Delta H_f$) is the enthalpy change when 1 mole of a substance is formed from its elements. If a substance is a reactant (being consumed), the enthalpy change for its formation in reverse (consumption) is the negative of its enthalpy of formation. But here, we are forming 2 moles of the intermediate as a reactant? Wait, no—wait, the problem says "2 moles of the intermediate are formed as a reactant"—wait, actually, if the intermediate is being formed (as a reactant? Wait, maybe wording: if the intermediate is a reactant, but we are forming it? Wait, no, enthalpy of formation is for formation. If we are forming the intermediate (even if it's a reactant in the next step), the enthalpy change for forming $n$ moles is $n \times \Delta H_f$. But wait, the sign: enthalpy of formation is positive if it's endothermic (absorbs heat) when forming. Wait, the enthalpy of formation of the intermediate is 90.3 kJ/mol, meaning forming 1 mole absorbs 90.3 kJ (endothermic). But if we are forming 2 moles, the enthalpy change for this step (forming 2 moles) would be $2 \times 90.3$? Wait, no—wait, the question is: the intermediate is formed as a reactant. Wait, maybe the intermediate is a reactant, but we are forming it (so the reaction step is forming the intermediate, which is a reactant in the overall reaction? Wait, no, the step is: during the reaction, 2 moles of the intermediate are formed as a reactant. Wait, maybe the reaction step is producing the intermediate (so the intermediate is a product of this step, but a reactant in the next). Wait, the enthalpy of formation is 90.3 kJ/mol for the intermediate. So forming 1 mole of intermediate has $\Delta H = +90.3$ kJ (since enthalpy of formation is given as 90.3 kJ/mol, meaning it's endothermic to form). So forming 2 moles: $\Delta H = 2 \times 90.3$ kJ? Wait, no—wait, the options: let's check. Wait, the options have 180.6 and -180.6. Wait, maybe the intermediate is a reactant, so the step is consuming the intermediate? No, the problem says "2 moles of the intermediate are formed as a reactant"—maybe wording is confusing. Wait, enthalpy of formation: when a substance is formed, $\Delta H_f$ is the enthalpy change. If the substance is a reactant (being broken down), the enthalpy change is $-\Delta H_f$ per mole. But here, the problem says "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." Wait, maybe "formed as a reactant" means that in this step, the intermediate is being formed (so it's a product of this step, but a reactant in the next), so the enthalpy change for forming 2 moles would be $2 \times 90.3$ kJ? But 90.3 * 2 = 180.6. But wait, the sign: enthalpy of formation is positive, so forming is endothermic (positive $\Delta H$). But let's check the options. Option B is 180.6 kJ, option C is -180.6. Wait, maybe I got the direction wrong. Wait, if the intermediate is a reactant, maybe the step is consuming the intermediate, but the problem says "formed as a reactant"—maybe it's a typo, and it's "formed as a product". Wait, no. Wait, let's re-express: Enthalpy of formation ($\Delta H_f$) is the enthalpy change when 1 mole is formed from elements. So if we form 2 moles, $\Delta H = 2 \times \Delta H_f$. But if the intermediate is a reactant, maybe the step is the reverse: breaking down the intermediate, but the problem says "formed". Wait, the problem says: "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?" So forming 2 moles: each mole has $\Delta H_f = 90.3$ kJ/mol, so total $\Delta H = 2 \times 90.3 = 180.6$ kJ. But wait, is forming a reactant endothermic or exothermic? Enthalpy of formation is positive, so forming is endothermic (requires heat), so $\Delta H$ is positive. So the answer should be 180.6 kJ, which is option B. Wait, but let's check again. Wait, maybe the intermediate is a reactant, so the step is consuming the intermediate, but the problem says "formed". Maybe the wording is that the intermediate is formed (produced) in this step, and then used as a reactant in the next. So in this step, forming 2 moles, so $\Delta H = 2 \times 90.3 = 180.6$ kJ. So option B.

Step2: Calculate total enthalpy change

Given $\Delta H_f = 90.3$ kJ/mol, moles $n = 2$.
Total enthalpy change $\Delta H = n \times \Delta H_f = 2 \times 90.3$ kJ.
$2 \times 90.3 = 180.6$ kJ.

Answer:

B. 180.6 kJ