The equilibrium constant is 0.0900 at 25^∘ C for the reaction H2 O(g)+Cl2 O(g) ⇌2 HOCl(g) For wh

The equilibrium constant is 0.0900 at 25^∘ C for the...

The equilibrium constant is 0.0900 at $25^{\circ} \mathrm{C}$ for the reaction $\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Cl}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HOCl}(g)$ For which of the following sets of conditions is the system at equilibrium? For those which are not at equilibrium, in which direction will the system shift? a. $P_{\mathrm{H}_{2} \mathrm{O}}=1.00 \mathrm{atm}, P_{\mathrm{C}_{2} \mathrm{O}}=1.00 \mathrm{atm}, P_{\mathrm{HOC}}=1.00 \mathrm{atm}$ b. $P_{\mathrm{H}_{2} \mathrm{O}}=200 .$ torr, $P_{\mathrm{C}_{2} \mathrm{O}}=49.8$ torr, $P_{\mathrm{HOCl}}=21.0$ torr c. $P_{\mathrm{H}_{2} \mathrm{O}}=296$ torr, $P_{\mathrm{C}_{2} \mathrm{O}}=15.0$ torr, $P_{\mathrm{HOC}}=20.0$ torr

The equilibrium constant is 0.0900 at $25^{\circ} \mathrm{C}$ for the reaction
$\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Cl}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HOCl}(g)$
For which of the following sets of conditions is the system at equilibrium? For those which are not at equilibrium, in which direction will the system shift?
a. $P_{\mathrm{H}_{2} \mathrm{O}}=1.00 \mathrm{atm}, P_{\mathrm{C}_{2} \mathrm{O}}=1.00 \mathrm{atm}, P_{\mathrm{HOC}}=1.00 \mathrm{atm}$
b. $P_{\mathrm{H}_{2} \mathrm{O}}=200 .$ torr, $P_{\mathrm{C}_{2} \mathrm{O}}=49.8$ torr, $P_{\mathrm{HOCl}}=21.0$ torr
c. $P_{\mathrm{H}_{2} \mathrm{O}}=296$ torr, $P_{\mathrm{C}_{2} \mathrm{O}}=15.0$ torr, $P_{\mathrm{HOC}}=20.0$ torr

Key Concepts

Equilibrium Constant

This is the ratio of the concentrations or partial pressures of the products to those of the reactants when a system is at equilibrium. For gas?phase reactions, the equilibrium constant (Kp) is expressed in terms of partial pressures. It is a fixed value at a given temperature, which indicates the extent of the reaction under equilibrium conditions.

Reaction Quotient

The reaction quotient (Q) is calculated using the same expression as the equilibrium constant but for conditions that are not necessarily at equilibrium. By comparing Q to K, one can determine the direction in which the reaction will shift to reach equilibrium; if Q < K, the reaction will shift to the right, and if Q > K, it will shift to the left.

Le Châtelier's Principle

Le Châtelier's Principle states that if a system at equilibrium is disturbed by a change in concentration, temperature, or pressure, the system will adjust to counteract the effect of the disturbance and re-establish equilibrium. This principle is used to predict the direction of the shift in equilibrium when conditions deviate from those at equilibrium.

Partial Pressures

In gas-phase reactions, the concentration of each gas is often expressed as its partial pressure. Partial pressures allow for the application of equilibrium expressions to gases, and they can be converted between different units (e.g., torr and atm) if necessary to compare with the equilibrium constant expressed in a specific unit.

Transcript

00:01 In this problem, we are looking at different conditions for the partial pressures of all of the gases in this equilibrium reaction.

00:13 We're going to figure out which of these equations in the following conditions are at equilibrium and which ones need to shift in a certain direction to achieve equilibrium.

00:25 So what we're going to be using in order to figure out how to solve this problem is q.

00:34 So q is a lot like our k constant, except q is calculated at any condition.

00:43 It does not have to be at equilibrium.

00:45 So in this case, we are using partial pressure.

00:48 So q is going to be equal to the partial pressure of the product, which is h -o -c -l raised to the power of its coefficient, so that's squared, and then divided by the product of the partial pressure of its first reactant, h -2 -l -w -w -water vapor, times the partial pressure of its second reactant, which is c -l -o -2.

01:28 So again, q is a lot like k, but it's not a, equilibrium necessarily, or it's at least not specified that we're calculating it at equilibrium.

01:43 But it does look a lot like k.

01:45 So we're going to be comparing it to k to figure out whether we're at equilibrium or not.

01:49 And we are actually given in the problem k, which is actually kp here, but it shouldn't matter.

01:59 The kp and the kc should be the same.

02:01 But that's only for this problem.

02:05 Since there are the same number of moles on both sides of the equation.

02:10 Kp and kc should be the same.

02:12 But in this case kp is equal to 0 .090.

02:18 So let's start plugging in our partial pressures for q.

02:24 So in part a we are given that's our first set of conditions.

02:31 We have one atmosphere of everything.

02:34 So we'll have one atm for water vapor, one for cl02 and then one for hocl.

02:57 So now let's plug in for q.

03:00 So q is going to be equal to 1 .00 squared divided by 1 times 1.

03:26 So it's going to be equal to point oh sorry not point.

03:33 It's going to be equal to 1 which is greater than i should do the right number of sigpigs so that's going to be 3 sigpigs 1 point zero and that is going to be greater than k which is point zero nine zero zero zero just k so if our q is larger than our k that means we have too much product so how do we know that well if we look at the equation for q we have the products partial pressure on top so when we have an excess amount of product it's going to have a higher partial pressure than it would at equilibrium.

04:20 And that's going to make our potion larger.

04:24 So in order to reverse that, we actually need to shift towards our reactants.

04:29 So when we shift towards our reactants, water vapor and clo2, when we do that, we lessen the amount of excess product, and we increase the amount of reactants.

04:40 So we're increasing the denominator, and we are reducing our numerator that's going to give us a smaller potion, eventually allowing us to reach k.

04:48 So in order to reach equilibrium, we need to shift towards the reactants.

04:55 And that's going to be shift left.

05:03 So we're going to be doing the exact same thing for the next two parts of this question...

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