Question

How much energy is required to decompose 612 g of pcl3, according to the reaction below? the molar mass of pcl3 is 137.32 g/mol and may be useful. 4 pcl3(g) ? p4(s) + 6 cl2(g) ?h°rxn = +1207 kj
a.1.85 × 103 kj

b.3.46 × 103 kj

c.5.38 × 103 kj

d.1.34 × 103 kj

e.4.76 × 103 kj?

Answer 1

The reaction is:

4 PCl3 (g) ---> P4(s) + 6 Cl2(g).

Now, you need to convert the mass of PCl3 into number of moles, for which you use the molar mass of PCl3 in this way:

number of moles = number of grams / molar mass =>

number of moles of PCl3 = 612 g / 137.32 g/mol = 4.4567 moles of PCl3.

Now use the proportion with the ΔH rxn given.

4 mol PCl3 / 1207 kJ = 4.4567 mol / x => x = 4.4567 mol * 1207 kJ / 4 mol = 1,344.8 kJ = 1.34 * 10^3 kJ.

Answer: 1.34 * 10 ^3 kJ (option d)

Answer 2

The energy required to decompose $612{\text{ g}}$ of ${\text{PC}}{{\text{l}}_3}$ is $\boxed{1.34 \times {{10}^3}{\text{ kJ}}}$.

Further Explanation:

Enthalpy:

It is a thermodynamic property that is defined as the sum of internal energy and product of pressure (P) and volume (V) of system. It is a state function, an extensive property, and is independent of the path followed by the system while moving from initial to the final point. The total enthalpy of the system cannot be measured directly so its change $\left( {\Delta H} \right)$ is usually measured.

The given chemical reaction is,

$4{\text{PC}}{{\text{l}}_3}\left( g\right)\to {{\text{P}}_4}\left( s \right) + 6{\text{C}}{{\text{l}}_2}\left( g \right)$               ......(1)

The value of $\Delta H_{{\text{rxn}}}^0$ for the above reaction is $+ 1207{\text{ kJ}}$.

First, calculate the number of moles of ${\text{PC}}{{\text{l}}_3}$ present in the reaction mixture.

The given mass of ${\text{PC}}{{\text{l}}_3}$ is $612{\text{ g}}$ and molar mass of ${\text{PC}}{{\text{l}}_3}$ is $137.32{\text{ g/mol}}$. The expression to calculate the number of moles of ${\text{PC}}{{\text{l}}_3}$ is as follows:

${\text{Number of moles of PC}}{{\text{l}}_3}=\dfrac{{{\text{Given mass of PC}}{{\text{l}}_3}}}{{{\text{Molar mass of PC}}{{\text{l}}_{\text{3}}}}}$                  

                                  ......(2)

Substitute $612{\text{ g}}$ for given mass of ${\text{PC}}{{\text{l}}_3}$, and $137.32{\text{ g/mol}}$ for molar mass of ${\text{PC}}{{\text{l}}_3}$ in equation (2).

$\begin{aligned}{\text{Number of moles of PC}}{{\text{l}}_3}&=\frac{{612{\text{ g}}}}{{137.32{\text{ g/mol}}}}\\&= 4.457{\text{ mol}}\\\end{aligned}$

According to the balance reaction, decomposition of 4 moles of ${\text{PC}}{{\text{l}}_3}$ requires $+ 1207{\text{ kJ}}$ energy, therefore, the energy required by 1 mole of ${\text{PC}}{{\text{l}}_3}$ can be calculated as follows:

$\begin{aligned}{\text{Energy required by 1 mol PC}}{{\text{l}}_3} &= \frac{{ + 1207{\text{ kJ}}}}{{4{\text{ mol}}}}\\&= 301.75{\text{ kJ/mol}}\\\end{aligned}$

The energy required by 4.46 mole of ${\text{PC}}{{\text{l}}_3}$ can be calculated as follows:

$\begin{aligned}{\text{Energy required by 4}}{\text{.46 mol PC}}{{\text{l}}_3}&=301.75{\text{ kJ/mol}} \times 4.457{\text{ mol}}\\&= 1344.9{\text{ kJ}}\\\end{aligned}$

Convert $1344.9{\text{ kJ}}$ into scientific notation as follows:

$\begin{aligned}{\text{Energy}} &= 1344.9{\text{ kJ}}\\ &\simeq 1.34 \times {10^3}{\text{ kJ}}\\\end{aligned}$

Hence the energy required to decompose $612{\text{ g}}$ of ${\text{PC}}{{\text{l}}_3}$ is $1.34 \times {10^3}{\text{ kJ}}$.

Learn more:

1.Calculate the enthalpy change using Hess’s Law:brainly.com/question/11293201

2. Find the enthalpy of decomposition of 1 mole of MgO:brainly.com/question/2416245

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Thermodynamics

Keywords: enthalpy, exothermic, endothermic, negative, positive, 1344.9 kJ, 1, moles, PCl3, P4, and Cl2.