Question

measurements show that the enthalpy of a mixture of gaseous reactants increases by 373. kj during a certain chemical reaction, which is carried out at a constant pressure. furthermore, by carefully monitoring the volume change it is determined that - 92. kj of work is done on the mixture during the reaction
calculate the change in energy of the gas mixture during the reaction. be sure your answer has the correct number of significant digits
kj
is the reaction exothermic or endothermic?
exothermic
endothermic

Explanation:

Step1: Recall the first law of thermodynamics

The first law of thermodynamics is given by the equation $\Delta U = q + w$, where $\Delta U$ is the change in internal energy, $q$ is the heat transferred, and $w$ is the work done. At constant pressure, the enthalpy change $\Delta H$ is equal to the heat transferred $q$ (i.e., $q = \Delta H$). Here, the enthalpy of the reactants increases by 373 kJ, so $q=\Delta H = 373\space kJ$ (since enthalpy increase means heat is absorbed, $q$ is positive). The work done on the mixture is $w=- 92\space kJ$? Wait, no: the sign convention: if work is done on the system, $w$ is positive; if work is done by the system, $w$ is negative. Wait, the problem says " - 92 kJ of work is done on the mixture". Wait, the wording: " - 92 kJ of work is done on the mixture" – maybe that's a typo? Wait, no, let's check the sign convention. The first law: $\Delta U=q + w$, where $q$ is heat added to the system, $w$ is work done on the system. So if the enthalpy increases by 373 kJ at constant pressure, that means $q = \Delta H=373\space kJ$ (because at constant pressure, $\Delta H = q_p$). The work done on the mixture is $w = - 92\space kJ$? Wait, no, the problem says " - 92 kJ of work is done on the mixture". Wait, maybe the work done on the system is $w = - 92\space kJ$? No, that doesn't make sense. Wait, maybe the work done by the system is - 92 kJ, so work done on the system is + 92 kJ? Wait, no, let's re - read: "by carefully monitoring the volume change it is determined that - 92 kJ of work is done on the mixture during the reaction". Wait, the sign: in the first law, $w$ is positive when work is done on the system, negative when work is done by the system. So if work is done on the mixture (system), $w$ should be positive. But the problem says " - 92 kJ of work is done on the mixture". That might be a mistake in the problem's wording, or maybe it's a sign error. Wait, maybe the work done by the system is 92 kJ, so work done on the system is $w=-92\space kJ$? No, that's confusing. Wait, let's go back. The enthalpy change $\Delta H = q_p=373\space kJ$ (since enthalpy of reactants increases, so heat is absorbed, $q$ is positive). Now, the work: if work is done on the system, $w$ is positive. But the problem says " - 92 kJ of work is done on the mixture". Wait, maybe that's a typo and it's supposed to be "92 kJ of work is done on the mixture" or " - 92 kJ of work is done by the mixture". Let's assume that the work done on the system is $w = - 92\space kJ$? No, that's not right. Wait, let's check the standard sign convention:

- $q>0$: heat is added to the system (endothermic)
- $q < 0$: heat is removed from the system (exothermic)
- $w>0$: work is done on the system
- $w < 0$: work is done by the system

The problem states: "the enthalpy of a mixture of gaseous reactants increases by 373 kJ during a certain chemical reaction, which is carried out at a constant pressure" – so at constant pressure, $\Delta H=q_p = 373\space kJ$ (since enthalpy increase means heat is absorbed, $q_p$ is positive). Then, "by carefully monitoring the volume change it is determined that - 92 kJ of work is done on the mixture during the reaction" – so work done on the system is $w=-92\space kJ$? That would mean work is done by the system (since $w$ is negative). Wait, maybe the problem has a sign error, and the work done on the system is $w = 92\space kJ$, but it's written as - 92 kJ. Alternatively, maybe the work done by the system is 92 kJ, so $w=-92\space kJ$. Let's proceed with the given values.

So $\Delta U=q + w$. We have $q=\Delta H = 373\space kJ$ (at constant pressure, $\Delta H = q_p$), and $w=-92\space kJ$ (since work is done on the mixture is - 92 kJ – maybe that means work is done by the mixture, so $w$ is negative). Then:

$\Delta U=373\space kJ+(- 92\space kJ)=373 - 92=281\space kJ$? Wait, no, that can't be. Wait, maybe the work done on the system is $w = 92\space kJ$, and the problem has a sign error (the - sign is misplaced). Let's re - evaluate the sign convention.

Wait, the enthalpy change: when the enthalpy of the reactants increases, that means the reaction is endothermic (since enthalpy of products is higher than reactants, so heat is absorbed). So $q=\Delta H = 373\space kJ$ (positive, because heat is absorbed by the system). Now, work done on the system: if the volume changes, for a gas, if the system is compressed, work is done on the system (positive $w$); if the system expands, work is done by the system (negative $w$). The problem says " - 92 kJ of work is done on the mixture" – maybe that's a mistake, and it's "92 kJ of work is done on the mixture", so $w = 92\space kJ$. Then $\Delta U=q + w=373+92 = 465\space kJ$? No, that doesn't match. Wait, no, let's check the first law again.

Wait, maybe I got the sign of $q$ wrong. Wait, enthalpy change: $\Delta H=H_{products}-H_{reactants}$. If the enthalpy of the reactants increases, that means $H_{products}>H_{reactants}$, so $\Delta H = H_{products}-H_{reactants}=373\space kJ$. At constant pressure, $\Delta H = q_p$, so $q_p = 373\space kJ$ (heat is absorbed by the system, so $q$ is positive). Now, work: the problem says " - 92 kJ of work is done on the mixture". So work done on the system is $w=-92\space kJ$ – which would mean work is done by the system (since $w$ is negative). So $\Delta U=q + w=373+(-92)=281\space kJ$. But let's check the significant figures. The values are 373 (three significant figures) and 92 (two significant figures). Wait, 373 has three, 92 has two. When adding or subtracting, the result should have the same number of decimal places as the least precise measurement. But 373 and 92 are whole numbers, so the result should be to the ones place. 373 - 92 = 281. Now, for the exothermic or endothermic: since the enthalpy increased (heat was absorbed), the reaction is endothermic.

Wait, maybe the work done on the system is $w = 92\space kJ$ (the - sign is a mistake). Let's recalculate: $\Delta U=373 + 92=465\space kJ$. But that contradicts the sign. Wait, no, the problem says " - 92 kJ of work is done on the mixture". So the work done on the system is - 92 kJ, meaning work is done by the system (so $w=-92\space kJ$). Then $\Delta U=q + w=373+(-92)=281\space kJ$.

Now, for the second part: since the enthalpy of the reactants increased (heat was absorbed by the system), the reaction is endothermic.

Step2: Calculate the change in internal energy

Using the first law of thermodynamics $\Delta U=q + w$. We know that at constant pressure, $q = \Delta H=373\space kJ$ (enthalpy increase means heat is absorbed, so $q$ is positive). The work done on the mixture is $w=-92\space kJ$ (as per the problem statement).

$\Delta U=373\space kJ+(-92\space kJ)=373 - 92 = 281\space kJ$

Step3: Determine if the reaction is exothermic or endothermic

Since the enthalpy of the reactants increased (heat was absorbed by the system), the reaction is endothermic. Exothermic reactions release heat (enthalpy of products is less than reactants), while endothermic reactions absorb heat (enthalpy of products is greater than reactants). Here, the enthalpy of the reactants (mixture) increased, so it's an endothermic reaction.

Answer:

The change in energy of the gas mixture is $\boldsymbol{281\space kJ}$, and the reaction is $\boldsymbol{endothermic}$.