Problem 109 - Chemistry Matter and Change

Question

Apply Phosphorus trichloride is a starting material for the preparation of organic phosphorous compounds. Demonstrate how thermochemical equations a and b can be used to determine the enthalpy change for the reaction \(\mathrm{PCl}_{3}(1)+\mathrm{Cl}_{2}(\mathrm{g}) \rightarrow \mathrm{PCl}_{5}(\mathrm{s})\) a. \(P_{4}(s)+6 C l_{2}(g) \rightarrow 4 P C l_{3}(1) \quad \Delta H=-1280 \mathrm{kJ}\) b. \(P_{4}(s)+10 C l_{2}(g) \rightarrow 4 P C l_{5}(s) \qquad \Delta H=-1774 \mathrm{kJ}\)

Step-by-Step Solution

Answer

The enthalpy change for the reaction \(PCl_3(l) + Cl_2(g) \rightarrow PCl_5(s)\) is \(\Delta H = -123.5 kJ\). This is determined by manipulating and combining the given thermochemical equations.

Step 1: Analyze provided thermochemical equations

We are given two thermochemical equations: Equation a: \(P_4(s) + 6Cl_2(g) \rightarrow 4PCl_3(l)\) with \(\Delta H_a=-1280 kJ\) Equation b: \(P_4(s) + 10Cl_2(g) \rightarrow 4PCl_5(s)\) with \(\Delta H_b=-1774 kJ\) Our goal is to determine the enthalpy change for the following reaction: \(PCl_3(l) + Cl_2(g) \rightarrow PCl_5(s)\)

Step 2: Manipulate the given thermochemical equations

From equation a, we want to isolate the PCl3 term. To do this, we will divide equation a by 4: \(\frac{1}{4}[P_4(s) + 6Cl_2(g) \rightarrow 4PCl_3(l)]\) \( \frac{1}{4}P_4(s) + \frac{3}{2}Cl_2(g) \rightarrow PCl_3(l)\) Since we divided the entire equation by 4, we must also divide its enthalpy change by 4: \(\Delta H'_a = \frac{1}{4} \Delta H_a = \frac{1}{4}(-1280 kJ) = -320 kJ\) We can now write our updated equation a: \( \frac{1}{4}P_4(s) + \frac{3}{2}Cl_2(g) \rightarrow PCl_3(l) \qquad \Delta H'_a=-320 kJ\) Now, we will manipulate equation b by dividing it by 4 to isolate the PCl5 term: \(\frac{1}{4}[P_4(s) + 10Cl_2(g) \rightarrow 4PCl_5(s)]\) \( \frac{1}{4}P_4(s) + \frac{5}{2}Cl_2(g) \rightarrow PCl_5(s)\) We must also divide its enthalpy change by 4: \(\Delta H'_b = \frac{1}{4} \Delta H_b = \frac{1}{4}(-1774 kJ) = -443.5 kJ\) Our updated equation b: \( \frac{1}{4}P_4(s) + \frac{5}{2}Cl_2(g) \rightarrow PCl_5(s) \qquad \Delta H'_b=-443.5 kJ\)

Step 3: Subtract updated equation a from updated equation b

We will now subtract equation a from equation b to get the desired reaction: \(( \frac{1}{4}P_4(s) + \frac{5}{2}Cl_2(g) \rightarrow PCl_5(s) ) - ( \frac{1}{4}P_4(s) + \frac{3}{2}Cl_2(g) \rightarrow PCl_3(l) )\) \(PCl_3(l) + Cl_2(g) \rightarrow PCl_5(s)\) Now, we will subtract the enthalpy change of the updated equation a from the updated equation b: \(\Delta H = \Delta H'_b - \Delta H'_a = (-443.5 kJ) - (-320 kJ) = -123.5 kJ\)

Step 4: Write the final enthalpy change for the desired reaction

We now have the enthalpy change for the desired reaction: \(PCl_3(l) + Cl_2(g) \rightarrow PCl_5(s) \qquad \Delta H=-123.5 kJ\)

Key Concepts

Enthalpy change

Enthalpy change is a fundamental concept in thermochemistry, focusing on heat energy changes during chemical reactions. Simply put, it represents the total energy absorbed or released when reactants are transformed into products.
For the reaction \[PCl_3(l) + Cl_2(g) \rightarrow PCl_5(s)\] we calculated an enthalpy change, \(\Delta H\), of -123.5 kJ. This negative sign tells us that the reaction is exothermic, meaning it releases heat to the surroundings.
Understanding enthalpy changes helps us predict whether reactions occur spontaneously or require external energy. It also aids in calculating energy efficiency in industrial chemical processes. To discern \(\Delta H\), the principle of Hess's law is often employed, which states that the total enthalpy change for a reaction is the same regardless of the path taken.

Phosphorus compounds

Phosphorus compounds, especially phosphorus trichloride (\(PCl_3\)) and phosphorus pentachloride (\(PCl_5\)), play a crucial role in chemistry. These compounds are often utilized as intermediates in synthesizing various organic phosphorus compounds.
Phosphorus is an essential element and is frequently found in its elemental form as \(P_4\), a white phosphorus molecule. When this element reacts with chlorine, it forms phosphorus trichloride and pentachloride, important for understanding chemical behavior in reactions and manufacturing processes.

• \(PCl_3\) is used in pesticides, flame retardants, and plasticizers.
• \(PCl_5\) finds its applications in the pharmaceutical sector and as a chlorinating agent.

Diverse applications of these compounds demonstrate their significance in industrial chemistry.

Chemical reactions

Chemical reactions involve the transformation of reactants into products and include energy change as a core component. For our specific exercise, we examine the reaction where phosphorus trichloride reacts with chlorine gas to form phosphorus pentachloride.
The transformation \[PCl_3(l) + Cl_2(g) \rightarrow PCl_5(s)\] showcases how one or more substances change chemically through bond breaking and formation, often hinted by the release or absorption of energy.In chemical reactions, various types such as synthesis, decomposition, single-displacement, and double-displacement can occur. Controlling these reactions, especially focusing on reactant conditions like temperature, concentration, and catalyst presence, plays a vital role in the yield and speed of the reaction.