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Consider the equilibrium$$2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)$$If nitrosyl bromide, NOBr, is 34 percent dissociated at $25^{\circ} \mathrm{C}$ and the total pressure is 0.25 atm, calculate $K_{P}$ and $K_{\mathrm{c}}$ for the dissociation at this temperature.

$$\begin{array}{l}K_{p}=9.5 \times 10^{-3} \\K_{c}=3.9 \times 10^{-4}\end{array}$$

Chemistry 102

Chapter 14

Chemical Equilibrium

Carleton College

Drexel University

University of Maryland - University College

University of Kentucky

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.

00:54

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.

07:20

Consider the equilibrium

08:31

06:31

Consider the equilibrium $…

02:56

A Nitrosyl bromide, NOBr, …

04:41

When 0.0322 mol of $\mathr…

06:12

When $0.0322$ mol of $\mat…

05:56

Consider the dissociation …

01:24

Calculate the value of the…

we need to calculate the values of K C as well as K P. For this reaction, given the information of the total pressure after 34% of the initial n o b. Our associates at 25 degrees Celsius, we need to set up a reaction table in order to keep track of the species concentrations. If we're given the total pressure, then we need to write out our reaction table in terms of partial pressures to help us first calculate the value of K P. And then we can use the equation that relates K, P and K C to find Casey. So we have some initial announce summon initial amount of NLB our guests initially present, and we can call that X and we don't have any of the products present initially. And then we know that 34% of that initial ext associates. So the change will be minus 0.34 times that initial amount, which we call X. Where is the initial partial pressure of that? An OBE or gas and so at equilibrium when we combine those terms? One x minus 0.34 x that comes out to 0.66 x and now for both of the products, we add that change for this to to to stoke yama tree. That change will will be the same and magnitude. But the sign will be different since we're producing ANO and not consuming it. So it's that same magnitude of the change of zero point 34 x and then based on this 2 to 1 story geometry, the magnitude of the change 0.34 x will be divided by two for the production of BR two. Guests certainly divides your 20.34 x by two. We are producing zero 0.17 x for the change of BR to and so the equilibrium expressions for the partial pressures of both of the products are here for 0.34 x and 0.17 x So these are the partial pressures of all species at equilibrium and were given the total equilibrium pressure which is equal to the sum of the personal pressures equilibrium for all gases species. So that's a partial pressure of an O B e r. Plus the partial pressure of no less a partial pressure of B R two. We're told that that total equilibrium pressure is equal to 0.25 atmospheres, and we add all of those equilibrium. Partial pressure expressions together 0.66 x corresponds to the equilibrium. Partial pressure of an OBE are we have that to 0.34 x Coreno and 0.17 x for B R. Two. Now when we combine all of those x terms on the right side of the equation that condenses down into 1.17 x and now we can divide 0.25 by 1.17 to solve for X. When we do, we see that X comes out to about 0.214 atmospheres and now to sulphur KP The equilibrium expression of that corresponds to based on the Sochi arm. A tree of the given reaction is the concentration in partial pressures. So the partial pressure of you know, guess squared times the partial pressure of br to gas divided by the partial pressure of an O. B. R squared and to find all those equilibrium, partial pressures, we plug in this value of X that we just found into each one of these three expressions. And so when we do that, we can plug in the values of those equilibrium partial pressures into our equilibrium expression. It's only plug in that value of X for the expression for the equilibrium. Partial pressure of an old gas. We see that it should come out to about zero point 0726 atmospheres. You square that term, he br two should come out to be about zero 0.0 363 atmospheres at equilibrium and piano. BRD equilibrium should come out to about zero point 141 atmospheres and we square that turn and I only plug that all in sulphur the ratio to determine the value of K p. We should get about 9.6 times 10 to the power of negative three. And now we need to solve for Casey. We know that K C is equal to KP. Times are t judy negative Delta and gas. So K C is equal to the value of K p that we just found which was 9.6 times 10 to the power of negative three are is the ideal gas constant 0.8 to 1 leaders times, atmospheres, mole times Kelvin He is a temperature 25 degrees Celsius. Corresponds to 298 Kelvin He raises to the power of negative delta. And guess we see that on the product side we have two plus one. So a total of three moles of gas and on the react inside we have a total of two moles of gas or three minus two inside delta and gas is equal to positive one. We take the negative of that. So we do our times t and raise it to the power negative one re multiplied by the value of K P that we saw for earlier. And when we plugged at all in, we should find that K C at this temperature comes out to about 3.9 times 10 Judy power of negative four

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