The initial concentrations or pressures of reactants and...

The initial concentrations or pressures of reactants and products are given for each of the following systems. Calculate the reaction quotient and determine the direction in which each system will proceed to reach equilibrium. (a) $2 \mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \quad K_{c}=17 ;\left[\mathrm{NH}_{3}\right]=0.20 M,\left[\mathrm{N}_{2}\right]=1.00 \mathrm{M},\left[\mathrm{H}_{2}\right]=1.00 \mathrm{M}$ (b) $2 \mathrm{NH}_{2}(g) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)$ $K_{P}=6.8 \times 10^{4} ; \mathrm{NH}_{3}=3.0 \mathrm{atm}, \mathrm{N}_{2}=2.0 \mathrm{atm}, \mathrm{H}_{2}=1.0 \mathrm{atm}$ (c) $2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \quad K_{c}=0.230 ;\left[\mathrm{SO}_{3}\right]=0.00 \mathrm{M},\left[\mathrm{SO}_{2}\right]=1.00 \mathrm{M},\left[\mathrm{O}_{2}\right]=1.00 \mathrm{M}$ (d) $2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \quad K_{P}=16.5 ; \mathrm{SO}_{3}=1.00 \mathrm{atm}, \mathrm{SO}_{2}=1.00 \mathrm{atm}, \mathrm{O}_{2}=1.00 \mathrm{am}$ (e) $2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{NOCl}(g) \quad K_{c}=4.6 \times 10^{4} ;[\mathrm{NO}]=1.00 \mathrm{M},\left[\mathrm{Cl}_{2}\right]=1.00 \mathrm{M},[\mathrm{NOCl}]=0 \mathrm{M}$ (f) $\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) \quad K_{P}=0.050 ; \mathrm{NO}=10.0 \mathrm{atm}, \mathrm{N}_{2}=\mathrm{O}_{2}=5 \mathrm{atm}$

The initial concentrations or pressures of reactants and products are given for each of the following systems. Calculate the reaction quotient and determine the direction in which each system will proceed to reach equilibrium.
(a) $2 \mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \quad K_{c}=17 ;\left[\mathrm{NH}_{3}\right]=0.20 M,\left[\mathrm{N}_{2}\right]=1.00 \mathrm{M},\left[\mathrm{H}_{2}\right]=1.00 \mathrm{M}$
(b) $2 \mathrm{NH}_{2}(g) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)$
$K_{P}=6.8 \times 10^{4} ; \mathrm{NH}_{3}=3.0 \mathrm{atm}, \mathrm{N}_{2}=2.0 \mathrm{atm}, \mathrm{H}_{2}=1.0 \mathrm{atm}$
(c) $2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \quad K_{c}=0.230 ;\left[\mathrm{SO}_{3}\right]=0.00 \mathrm{M},\left[\mathrm{SO}_{2}\right]=1.00 \mathrm{M},\left[\mathrm{O}_{2}\right]=1.00 \mathrm{M}$
(d) $2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \quad K_{P}=16.5 ; \mathrm{SO}_{3}=1.00 \mathrm{atm}, \mathrm{SO}_{2}=1.00 \mathrm{atm}, \mathrm{O}_{2}=1.00 \mathrm{am}$
(e) $2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{NOCl}(g) \quad K_{c}=4.6 \times 10^{4} ;[\mathrm{NO}]=1.00 \mathrm{M},\left[\mathrm{Cl}_{2}\right]=1.00 \mathrm{M},[\mathrm{NOCl}]=0 \mathrm{M}$
(f) $\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) \quad K_{P}=0.050 ; \mathrm{NO}=10.0 \mathrm{atm}, \mathrm{N}_{2}=\mathrm{O}_{2}=5 \mathrm{atm}$

Key Concepts

Using Concentrations vs. Partial Pressures

In equilibrium problems, the form of the equilibrium constant expression depends on whether the system is described in terms of molar concentrations (Kc) or partial pressures (Kp). This distinction is crucial because the units and numerical values differ based on the phase of the reactants and products, and proper use of either Kc or Kp ensures consistency when calculating Q and comparing it with K.

Comparing Q and K

By comparing the reaction quotient (Q) to the equilibrium constant (K), one can predict the direction in which a reaction will shift to achieve equilibrium. If Q is less than K, the reaction will proceed in the forward direction, producing more products. Conversely, if Q is greater than K, the reaction will shift in the reverse direction, forming more reactants. When Q equals K, the system is already at equilibrium.

Stoichiometry in Equilibrium Expressions

Accurate formulation of both Q and K relies on the balanced chemical equation, where the concentrations or pressures of the reactants and products are raised to the power of their coefficients. This reflects the stoichiometric relationship in which reactants combine and products form, ensuring that the mathematical expressions appropriately represent the reaction dynamics.

Reaction Quotient (Q)

The reaction quotient, Q, is a ratio that compares the current concentrations (or pressures) of the products and reactants of a reaction to their stoichiometric coefficients. It is calculated at any point in time, not just at equilibrium, and expressed similarly to the equilibrium constant. This concept allows one to assess how far a system is from reaching equilibrium.

Equilibrium Constant (K)

The equilibrium constant, K, quantifies the ratio of product to reactant concentrations (or pressures) when a reaction has reached a state of equilibrium. It is specific to a given reaction at a defined temperature, and its numerical value indicates the extent to which a reaction favors products or reactants under equilibrium conditions.

Transcript

00:01 Okay, so this is the equilibrium question.

00:09 This question asks us to calculate the reaction quotient and determine the direction in which each system will proceed to reach the equilibrium.

00:21 So before we solve this problem, let's first review.

00:25 How can we calculate the reaction quotient? basically, for any reaction, you have a, plus b and we will produce c and d.

00:52 And the reaction coattent were actually equals to, right? the concentration of the c power q times the concentration of d power p over the concentration of a, power n, times the concentration of b power m.

01:21 If in the reaction of the reactant or product are gases, you can also use pressure to represent.

01:33 And it's actually pretty similar.

01:35 You just use pressure c power q times the pressure d power to p and then the pressure a power n, pressure a power n, and the pressure b power m.

02:06 And this is how we calculate the reaction quotient.

02:10 As for to determine how the system will proceed to reach the equilibrium, basically we just compare the q, the reaction quotient with the equilibrium constant.

02:27 If the q is smaller than the k, direction will actually go forward to reach the equilibrium.

02:38 If the reaction quotient is larger than k, then direction will actually move back forward, backward direction to reach the equilibrium.

02:50 And if q equals to k, it's in the equilibrium state.

02:56 So the concentration of the system of the reactant or product were not changed.

03:07 So now let's work on the questions.

03:13 So the first chemical reaction is ammonia decompose to become nitrogen and sweet hydrogen gas.

03:39 And it already gives the reaction equilibrium constant, which is 17.

03:48 So now what i need to do is to calculate the reaction coattient.

03:54 And in this problem, it gives us the concentration, so we can express the reaction coattient with the concentration.

04:05 So the reaction coitent, which is equal to the concentration, of the nitrogen times the concentration of the hydrogen power three because there's a sweet here right and over the concentration of ammonium which should power two and now we can plug in the numbers the concentration of nitrogen is actually one more and the concentration of the hydrogen is also one more times 3, power 3, and the concentration of the ammonium is 0 .2 more, power 2.

05:06 And it will actually equal to 1 more times 4, power 4, over 0 .04, more, 4, more 0 .04, more .4, more.

05:20 2.

05:23 So the answer is actually 25 more square.

05:34 And this number is larger than the kc, the equilibrium constants, which means direction will proceed backwards, the left direction.

05:56 Okay, now we can try to look as the second question, which is the same reaction.

06:11 Ammonian, decompose, become nitrogen, and hydrogen gas.

06:25 But this time, the equation constant are represented with pressure, the kp, which is 6 .8 times 10 to 4.

06:40 And also it gives us the pressure of each gas.

06:48 So here we need to use a partial pressure to represent the reaction coattients.

06:56 And we can work on that.

07:00 So the pressure version of the reaction coitient, which is equal to the pressure of the nitrogen times the pressure of the hydrogen, power 3 over the pressure of the ammonia, power 2.

07:22 And then we can plug in the numbers.

07:27 The pressure of the nitrogen is around 2 atm.

07:39 And the pressure of the hydrogen is 1 atm, power 3, and the pressure of the ammonium is 380m, and you'll find the answer here, which is equal to 0 .22 .80m square.

08:33 And this is actually much smaller than the kp, which means the reaction will actually proceed forward, which is in the right direction.

09:07 And now let's look at the third one...

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