Respuesta :
The amount of energy required to decompose 800 g of [tex]{\text{PC}}{{\text{l}}_{\text{3}}}[/tex] is [tex]\boxed{{\text{1757}}{\text{.8 kJ}}}[/tex].
Further explanation:
Decomposition reactions:
The opposite of combination reactions is called a decomposition reaction. Here, a single reactant gets broken into two or more products. Such reactions are usually endothermic because energy is required to break the existing bonds between the reactant molecules.
Examples of decomposition reactions are as follows:
(a) [tex]2{{\text{H}}_2}{{\text{O}}_2}\to2{{\text{H}}_2}{\text{O}}+{{\text{O}}_2}[/tex]
(b) [tex]2{\text{NaCl}}\to{\text{2Na + C}}{{\text{l}}_2}[/tex]
Stoichiometry of a reaction is used to determine the amount of species present in the reaction by the relationship between reactants and products. It is used to determine moles of a chemical species when moles of other chemical species present in the reaction is given.
Consider the general reaction,
[tex]{\text{A}}+2{\text{B}}\to3{\text{C}}[/tex]
Here,
A and B are reactants.
C is the product.
One mole of A reacts with two moles of B to produce three moles of C. The stoichiometric ratio between A and B is 1:2, the stoichiometric ratio between A and C is 1:3 and the stoichiometric ratio between B and C is 2:3.
The decomposition of [tex]{\text{PC}{\text{l}_3}[/tex] occurs as follows:
[tex]{\text{4PC}}{{\text{l}}_3} \to {{\text{P}}_4} + 6{\text{C}}{{\text{l}}_2}[/tex]
The formula to calculate the moles of [tex]{\text{PC}}{{\text{l}}_3}[/tex] is as follows:
[tex]{\text{Moles of PC}}{{\text{l}}_3}=\frac{{{\text{Given mass of PC}}{{\text{l}}_3}}}{{{\text{Molar mass of PC}}{{\text{l}}_3}}}[/tex] …… (1)
The given mass of [tex]{\text{PC}}{{\text{l}}_3}[/tex]is 800 g.
The molar mass of [tex]{\text{PC}}{{\text{l}}_3}[/tex]is 137.33 g/mol.
Substitute these values in equation (1).
[tex]\begin{aligned}{\text{Moles of PC}}{{\text{l}}_3}&=\left( {{\text{800 g}}}\right)\left({\frac{{{\text{1 mol}}}}{{{\text{137}}{\text{.33 g}}}}}\right)\\&=5.8253\;{\text{mol}}\\\end{aligned}[/tex]
From the balanced chemical reaction of decomposition of [tex]{\text{PC}}{{\text{l}}_3}[/tex], 4 moles of [tex]{\text{PC}}{{\text{l}}_3}[/tex] require 1207 kJ of energy. So the amount of energy required to decompose 5.8253 moles of [tex]{\text{PC}}{{\text{l}}_3}[/tex] is calculated as follows:
[tex]\begin{aligned}{\text{Amount of energy required}}&=\left({5.8253\;{\text{mol PC}}{{\text{l}}_3}}\right)\left({\frac{{{\text{1207 kJ}}}}{{{\text{4 mol PC}}{{\text{l}}_3}}}}\right)\\&={\text{1757}}{\text{.7842}}\;{\text{kJ}}\\&\approx{\text{1757}}{\text{.8 kJ}}\\\end{aligned}[/tex]
So the amount of energy required to decompose 800 g of [tex]{\mathbf{PC}}{{\mathbf{l}}_{\mathbf{3}}}[/tex] is 1757.8 kJ.
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Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Mole Concept
Keywords: PCl3, P4, Cl2, 6 Cl2, 4 PCl3, amount of energy, 1757.8 kJ, moles of PCl3, molar mass, given mass, 800 g, 137.33 g/mol, stoichiometry, reactant, product, decomposition reaction, 1207 kJ.
To determine the quantity of energy required to decompose 800 grams of [tex]PCl_3[/tex] is 1,757.82 kJ.
Given the following data:
- Mass of [tex]PCl_3[/tex] = 800 grams
- Molar mass of [tex]PCl_3[/tex] = 137.32 g/mol
To determine the quantity of energy required to decompose 800 grams of [tex]PCl_3[/tex]:
First of all, we would write the properly balanced chemical equation for the decomposition of [tex]PCl_3[/tex]:
[tex]4PCl_3_{(g)} --->P_4_{(s)} +6Cl_2_{(g)} \ \;\delta h^{\circ}rxn =+1207 \;kJ[/tex]
Next, we would find the number of moles of [tex]PCl_3[/tex] contained in the chemical equation:
[tex]Number\;of\;moles = \frac{Mass}{molar\;mass}\\\\Number\;of\;moles = \frac{800}{137.32 }[/tex]
Number of moles = 5.8254 moles.
By stoichiometry:
4 moles of [tex]PCl_3[/tex] = 1207 kJ
5.8254 moles of [tex]PCl_3[/tex] = X kJ
Cross-multiplying, we have:
[tex]4 \times X = 1207 \times 5.8254\\\\4X=7,031.2578\\\\X=\frac{7,031.2578}{4}[/tex]
Energy, X = 1,757.82 kJ
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