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At equilibrium, the pressure of the reacting mixture$$\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$$is 0.105 atm at $350^{\circ} \mathrm{C}$. Calculate $K_{P}$ and $K_{\mathrm{c}}$ for this reaction.

$$\begin{array}{l}K_{p}=0.105 \\K_{c}=2.05 \times 10^{-3}\end{array}$$

Chemistry 102

Chapter 14

Chemical Equilibrium

Carleton College

Drexel University

University of Maryland - University College

Lectures

10:03

In thermodynamics, a state of thermodynamic equilibrium is a state in which a system is in thermal equilibrium with its surroundings. A system in thermodynamic equilibrium is in thermal equilibrium, mechanical equilibrium, electrical equilibrium, and chemical equilibrium. A system is in equilibrium when it is in thermal equilibrium with its surroundings.

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In chemistry, chemical equilibrium (also known as dynamic equilibrium) is a state of chemical stability in which the concentrations of the chemical substances do not change in the course of time due to their reaction with each other in a closed system. Chemical equilibrium is an example of dynamic equilibrium, a thermodynamic concept.

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for the following equation were given the values off 0.1058 p. M for the pressure and 350 degrees Celsius for the temperature. Since there is only one substance in the gaseous form just carbon dioxide, we know that carbon dioxide is the Onley substance that is contributing to the pressure. Because of this, we know the KP is equal zero point one 05 So now that we have the KP value, we're now able to find Casey using the following equation. Looking closer at what each component of this equation means we know our are is a constant is the ideal gas constant. And these are the standard units for the ideal gas constant. We know that t is the absolute temperature which we were given to be 350 degree Celsius. But we want this value in Calvin so that we have the same units carried throughout the equation. So that means converted. This is 600 23 degrees Kelvin, Did you find Delta and subscript Gene? You take all the moles of the product in the gaseous form, minus all the moles of the reactors in the gaseous form Looking at the products first, we have one substance, carbon dioxide that's in the gaseous form. And we have understood one mole here. And since we have no gaseous substances on the reaction side, we put zero so Delta and so screwed. G has a value off one. It's not that we have all the values we Dean. We can plug it into the equation. You know that 0.105 is equal to whatever Casey is, which is what we're finding. Times 0.0 8 to 1. There's a one here. Sorry times the temperature, which we found to be 623 raised to the one power. And if we are to isolate Casey, we see than it 0.105 Divided by ideal gas, constant times the absolute temperature in Calvin and solved. This is equivalent to 2.5 times 10 to the negative three

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