Tipo de Letra (3) Suppose that instead of adding N2, as in (2), you increase the concentration of Hz from 1.00 M (in #1) to 2.00 M, leaving [N2] = 1.00 M and [NH3] = 4.00 M. Is the system at equilibrium now? If not, which way will the reaction go? First: What would LeChatelier's Principle say? %3D To prove your answer, calculate the new reaction quotient, Qc. Is Qcgreater than, less than, or equal to Kc? (Circle one). If Qc is not equal to Kc, what would have to happen to make it equal 16? (4) Can you make a general statement about what happens to a reaction at equilibrium when you add either reactant? How would this be different for a reaction that "goes to completion"?

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O Procurar
Equilibrium v 2021 KS..
-quema
Referências
Correio
Rever
Ver
Ajuda
A A Aa A E-E-E
三三A T
AaBbCeDdl aBbCcD AaRbCcbdFe AABE
- 12
AaBbCcDdE
Ênfase
Forte
T Legenda
1 Normal
Subti
b x, x' A- eA-
Estilos
Parágrafo
Tipo de Letra
This set of questions addresses what would happen if a system at equilibrium is disturbed in different ways.
Consider the reaction:
N2(g) + 3 H2(g) → 2 NH3 (g)
[NH,
When the reaction is at equilibrium, the ratio of concentrations,
is CONSTANT and this constant
is called Kc. For this entire set of questions, take the value of Kc for this reaction to be 16.
[NH
When the reaction is NOT at equilibrium, the ratio
[N,][H,J
can take on just about any value.
For example, if we start with all reactants and zero products, then the ratio would equal zero. When not at
equilibrium, we call this ratio Qc. It is referred to as the "reaction quotient".
So, in summary:
Qc can be any value
when the reaction is not
When the reaction is at equilibrium K, -
[NH;]
- = 16.
at equilibrium.
The value of Qc just
depends on the
[NH,]
[N J[H,
When the reaction is NOT at equilibrium, Qc =
#Constant
concentrations of the R's
and P's in the reaction
(1) If the concentrations of the reactants and products are as follows:
[N2] = 1.00 M, [H2] = 1.00 M, [NH3] = 4.00 M .. is this system at equilibrium?
Prove your answer by figuring out the value of Qc and seeing if it equals 16. If it
does, the reaction is at equilibrium.
mixture.
(2) Suppose you increase the concentration of N2 from 1.00M to 2.00 M, leaving
[H2] =1.00 M and [NH3] = 4.00 M. Is the system at equilibrium now? If not, which way will the reaction go?
First: What would LeChatelier's Principle say?
To prove your answer, calculate the new reaction quotient, Qc.
Is Qcgreater than, less than, or equal to Kc? (Circle one).
If Qc is not equal to Kc, what would have to happen to make it equal 16?
stados Unidos)
Concentração
Transcribed Image Text:O Procurar Equilibrium v 2021 KS.. -quema Referências Correio Rever Ver Ajuda A A Aa A E-E-E 三三A T AaBbCeDdl aBbCcD AaRbCcbdFe AABE - 12 AaBbCcDdE Ênfase Forte T Legenda 1 Normal Subti b x, x' A- eA- Estilos Parágrafo Tipo de Letra This set of questions addresses what would happen if a system at equilibrium is disturbed in different ways. Consider the reaction: N2(g) + 3 H2(g) → 2 NH3 (g) [NH, When the reaction is at equilibrium, the ratio of concentrations, is CONSTANT and this constant is called Kc. For this entire set of questions, take the value of Kc for this reaction to be 16. [NH When the reaction is NOT at equilibrium, the ratio [N,][H,J can take on just about any value. For example, if we start with all reactants and zero products, then the ratio would equal zero. When not at equilibrium, we call this ratio Qc. It is referred to as the "reaction quotient". So, in summary: Qc can be any value when the reaction is not When the reaction is at equilibrium K, - [NH;] - = 16. at equilibrium. The value of Qc just depends on the [NH,] [N J[H, When the reaction is NOT at equilibrium, Qc = #Constant concentrations of the R's and P's in the reaction (1) If the concentrations of the reactants and products are as follows: [N2] = 1.00 M, [H2] = 1.00 M, [NH3] = 4.00 M .. is this system at equilibrium? Prove your answer by figuring out the value of Qc and seeing if it equals 16. If it does, the reaction is at equilibrium. mixture. (2) Suppose you increase the concentration of N2 from 1.00M to 2.00 M, leaving [H2] =1.00 M and [NH3] = 4.00 M. Is the system at equilibrium now? If not, which way will the reaction go? First: What would LeChatelier's Principle say? To prove your answer, calculate the new reaction quotient, Qc. Is Qcgreater than, less than, or equal to Kc? (Circle one). If Qc is not equal to Kc, what would have to happen to make it equal 16? stados Unidos) Concentração
ESLITOS
Parágrafo
Tipo de Letra
(3) Suppose that instead of adding N2, as in (2), you increase the concentration of H2 from 1.00 M (in #1) to
2.00 M, leaving [N2] = 1.00 M and [NH3] = 4.00 M. Is the system at equilibrium now? If not, which way will
the reaction go? First: What would LeChatelier's Principle say?
To prove your answer, calculate the new reaction quotient, Qc.
Is Qcgreater than, less than, or equal to Kc? (Circle one).
If Qc is not equal to Kc, what would have to happen to make it equal 16?
(4) Can you make a general statement about what happens to a reaction at equilibrium when you add either
reactant? How would this be different for a reaction that "goes to completion"?
13
(5) In the lab, we added a great excess of one reactant and stated that this would make the reaction go
almost all the way to the right (to completion). This question uses the above reaction to explore this
phenomenon. Suppose you start with the reactants at equilibrium with concentrations as in (#1):
[N2] = 1.00 M, [H2] = 1.00 M, [NH3] = 4.00M
DConcentração
idos)
Transcribed Image Text:ESLITOS Parágrafo Tipo de Letra (3) Suppose that instead of adding N2, as in (2), you increase the concentration of H2 from 1.00 M (in #1) to 2.00 M, leaving [N2] = 1.00 M and [NH3] = 4.00 M. Is the system at equilibrium now? If not, which way will the reaction go? First: What would LeChatelier's Principle say? To prove your answer, calculate the new reaction quotient, Qc. Is Qcgreater than, less than, or equal to Kc? (Circle one). If Qc is not equal to Kc, what would have to happen to make it equal 16? (4) Can you make a general statement about what happens to a reaction at equilibrium when you add either reactant? How would this be different for a reaction that "goes to completion"? 13 (5) In the lab, we added a great excess of one reactant and stated that this would make the reaction go almost all the way to the right (to completion). This question uses the above reaction to explore this phenomenon. Suppose you start with the reactants at equilibrium with concentrations as in (#1): [N2] = 1.00 M, [H2] = 1.00 M, [NH3] = 4.00M DConcentração idos)
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