Question

In the preparation of $CaO$ from $CaC{O_3}$ using the equilibrium, $CaC{O_3}\left( s \right){\text{ }} < \- - > {\text{ }}CaO{\text{ }}\left( s \right){\text{ }} + {\text{ }}C{O_2}\left( g \right)$
$Log{\text{ }}Kp{\text{ }} = {\text{ }}7.282{\text{ }}-{\text{ }}8500/{\text{ }}T$ for complete decomposition of $CaC{O_3}$, the temperature in Celsius?
A) $1167$
B) $894$
C) $8500$
D) $850$

Answer 1

Temperature is one of the important parameters in chemistry. There are three units for measuring the temperature. There are degrees Celsius, kelvin, and Fahrenheit. Exothermic and endothermic reactions are based on temperature. If the temperature of the reaction decreases, that reaction is called an endothermic reaction. If the temperature of the reaction increases, that reaction is called an exothermic reaction.
Formula used:
The measuring purpose of the temperature is so many scales are available. There are degrees Celsius scale, Kelvin scale, and Fahrenheit scale.
The formula for convert degree Celsius to kelvin in temperature
${\text{kelvin = degree + 273}}$
The formula for convert degree Kelvin to Celsius in temperature
${\text{Degree = kelvin - 273}}$

Complete answer:
The given data is below,
In the preparation of $CaO$ from $CaC{O_3}$ using the equilibrium, $CaC{O_3}\left( s \right){\text{ }} < \- - > {\text{ }}CaO{\text{ }}\left( s \right){\text{ }} + {\text{ }}C{O_2}\left( g \right)$
$Log{\text{ }}Kp{\text{ }} = {\text{ }}7.282{\text{ }}-{\text{ }}8500/{\text{ }}T$for complete decomposition of $CaC{O_3}$.
The above-mentioned chemical reaction is an endothermic reaction. Because $C{O_2}$ is released in the chemical reaction.
The Kp value of the above reaction is one.
$Kp = 1$
$Log\left( 1 \right){\text{ }} = {\text{ }}0$

$$Log{\text{ }}Kp{\text{ }} = {\text{ }}7.282{\text{ }}-{\text{ }}\dfrac{{8500}}{T} \\\ {\text{0 }} = {\text{ }}7.282{\text{ }}-{\text{ }}\dfrac{{8500}}{T} \\\ 7.282{\text{ = }}\dfrac{{8500}}{T} \\\ T{\text{ = }}\dfrac{{8500}}{{7.282}} \\\ T = 1167K \\\$$

The formula for convert degree Kelvin to Celsius in temperature

$${\text{Degree = kelvin - 273}} \\\ {\text{ = }}1167K - 273 \\\ = 894 \\\$$

According to the above discussion, we conclude in the preparation of $CaO$ from $CaC{O_3}$ using the equilibrium, CaC{O_3}\left( s \right){\text{ }} < \- - > {\text{ }}CaO{\text{ }}\left( s \right){\text{ }} + {\text{ }}C{O_2}\left( g \right)$$$$Log{\text{ }}Kp{\text{ }} = {\text{ }}7.282{\text{ }}-{\text{ }}8500/{\text{ }}Tfor complete decomposition of $CaC{O_3}$, the temperature in $894$°C.

Hence, option B is the correct answer.

Note:
In thermodynamics, two types of the main reaction are considered. One is an exothermic reaction and the other is an endothermic reaction. Exothermic reaction means heat flow from system to surround, heat flow from reaction mass to the environment. Endothermic reaction means heat flow from surrounding to system, heat flow from the environment to the reaction mass. Simply we can say endothermic reaction means heat absorbed in the reaction mass and exothermic reaction means heat evolved in the reaction mass. In chemistry thermodynamics is one of the important topics for monitoring the reaction. In thermodynamics, the word meaning is thermos means heat, and dynamic means flow. In thermodynamics, we study the flow of the heat in the chemical reaction. In this topic, we focus on three main things. There are systems, surroundings, and boundaries. We consider our reaction as a system and the environment is surrounding and in between the junction is the boundary.