You wish to know the standard formation enthalpy of liquid PCl3 P4( s)+6 Cl2( g) ⟶4 PCl3(ℓ) Thes

Name: Problem 74, Chapter 4: Chemistry The Molecular Science
Uploaded: 2021-07-10T15:20:44-08:00
Duration: 2 min 22 s
Channel: Lottie Adams
Description: You wish to know the standard formation enthalpy of liquid PCl3 P4( s)+6 Cl2( g) ⟶4 PCl3(ℓ) These reaction enthalpies have been determined experimentally: P4( s)+10 Cl2( g) ⟶ 4 PCl5( s) Δr H^∘=-1774.0 kJ / mol PCl3(ℓ)+Cl2( g) ⟶PCl5( s) Δr H^∘=-123.8 kJ / mol Calculate the formation enthalpy for PCl3(ℓ).

step 1 So here we're given two different reactions. And because we have PCL3 in the product side in the

Question

You wish to know the standard formation enthalpy of liquid $\mathrm{PCl}_{3}$ $$ \mathrm{P}_{4}(\mathrm{~s})+6 \mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow 4 \mathrm{PCl}_{3}(\ell) $$ These reaction enthalpies have been determined experimentally: $$ \begin{array}{ll} \mathrm{P}_{4}(\mathrm{~s})+10 \mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow 4 \mathrm{PCl}_{5}(\mathrm{~s}) & \Delta_{\mathrm{r}} H^{\circ}=-1774.0 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{PCl}_{3}(\ell)+\mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow \mathrm{PCl}_{5}(\mathrm{~s}) & \Delta_{\mathrm{r}} H^{\circ}=-123.8 \mathrm{~kJ} / \mathrm{mol} \end{array} $$ Calculate the formation enthalpy for $\mathrm{PCl}_{3}(\ell)$.

   You wish to know the standard formation enthalpy of liquid $\mathrm{PCl}_{3}$
$$
\mathrm{P}_{4}(\mathrm{~s})+6 \mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow 4 \mathrm{PCl}_{3}(\ell)
$$
These reaction enthalpies have been determined experimentally:
$$
\begin{array}{ll}
\mathrm{P}_{4}(\mathrm{~s})+10 \mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow 4 \mathrm{PCl}_{5}(\mathrm{~s}) & \Delta_{\mathrm{r}} H^{\circ}=-1774.0 \mathrm{~kJ} / \mathrm{mol} \\
\mathrm{PCl}_{3}(\ell)+\mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow \mathrm{PCl}_{5}(\mathrm{~s}) & \Delta_{\mathrm{r}} H^{\circ}=-123.8 \mathrm{~kJ} / \mathrm{mol}
\end{array}
$$
Calculate the formation enthalpy for $\mathrm{PCl}_{3}(\ell)$.

Chemistry The Molecular Science

John W. Moore,… 5th Edition

Chapter 4, Problem 74

Instant Answer

Solved by Expert Lottie Adams

07/10/2021

Step 1

8 \mathrm{~kJ} / \mathrm{mol} $$ We need to reverse this reaction to get $\mathrm{PCl}_{3}(\ell)$ on the product side, which gives us: $$ \mathrm{PCl}_{5}(\mathrm{~s}) \longrightarrow \mathrm{PCl}_{3}(\ell)+\mathrm{Cl}_{2}(\mathrm{~g}) \quad \Delta_{\mathrm{r}} Show more…

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You wish to know the standard formation enthalpy of liquid $\mathrm{PCl}_{3}$ $$ \mathrm{P}_{4}(\mathrm{~s})+6 \mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow 4 \mathrm{PCl}_{3}(\ell) $$ These reaction enthalpies have been determined experimentally: $$ \begin{array}{ll} \mathrm{P}_{4}(\mathrm{~s})+10 \mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow 4 \mathrm{PCl}_{5}(\mathrm{~s}) & \Delta_{\mathrm{r}} H^{\circ}=-1774.0 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{PCl}_{3}(\ell)+\mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow \mathrm{PCl}_{5}(\mathrm{~s}) & \Delta_{\mathrm{r}} H^{\circ}=-123.8 \mathrm{~kJ} / \mathrm{mol} \end{array} $$ Calculate the formation enthalpy for $\mathrm{PCl}_{3}(\ell)$.

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Key Concepts

Standard Enthalpy of Formation

The standard enthalpy of formation is the heat change that results when one mole of a compound is formed from its elements in their most stable forms under standard conditions. It serves as a benchmark for assessing the energetics of chemical reactions.

Hess’s Law

Hess’s Law states that the total enthalpy change for a reaction is the same regardless of the number of steps taken to complete the reaction. This principle allows us to calculate unknown reaction enthalpies by algebraically combining multiple reactions with known enthalpy changes.

Reaction Enthalpy

Reaction enthalpy is the measure of heat absorbed or released during a chemical reaction at a constant pressure. It is an important concept in thermodynamics used to evaluate the energy changes that accompany chemical transformations.

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