question 20

for the first reaction:
(\boldsymbol{ ext{maroon molecule}}
ightleftharpoons \boldsymbol{ ext{blue molecule}}) (k = 4)
at equilibrium?
(circ) yes (circ) no

for the second reaction:
(\boldsymbol{ ext{gray molecule}}
ightleftharpoons \boldsymbol{ ext{purple molecule}}) (k=\frac{1}{5})
at equilibrium?
(circ) yes (circ) no

for the third reaction:
(\boldsymbol{ ext{light blue molecule}}
ightleftharpoons \boldsymbol{ ext{lavender molecule}}) (k = \frac{2}{3})
at equilibrium?
(circ) yes (circ) no

for the fourth reaction:
(\boldsymbol{ ext{teal molecule (3 - atom)}}
ightleftharpoons \boldsymbol{ ext{beige molecule (3 - atom)}}) (k=\frac{1}{3})
at equilibrium?
(circ) yes (circ) no

Question

question 20

for the first reaction:
(\boldsymbol{	ext{maroon molecule}}
ightleftharpoons \boldsymbol{	ext{blue molecule}}) (k = 4)
at equilibrium?
(circ) yes (circ) no

for the second reaction:
(\boldsymbol{	ext{gray molecule}}
ightleftharpoons \boldsymbol{	ext{purple molecule}}) (k=\frac{1}{5})
at equilibrium?
(circ) yes (circ) no

for the third reaction:
(\boldsymbol{	ext{light blue molecule}}
ightleftharpoons \boldsymbol{	ext{lavender molecule}}) (k = \frac{2}{3})
at equilibrium?
(circ) yes (circ) no

for the fourth reaction:
(\boldsymbol{	ext{teal molecule (3 - atom)}}
ightleftharpoons \boldsymbol{	ext{beige molecule (3 - atom)}}) (k=\frac{1}{3})
at equilibrium?
(circ) yes (circ) no

Accepted Answer

### First Equilibrium (Brown ⇌ Blue, $ K = 4 $)
# Explanation:
## Step 1: Count Moles
Count brown (reactant) and blue (product) molecules. Brown: 2, Blue: 8.
## Step 2: Calculate Reaction Quotient ($ Q $)
For reaction $ \text{Brown}
ightleftharpoons \text{Blue} $, $ Q = \frac{[\text{Blue}]}{[\text{Brown}]} = \frac{8}{2} = 4 $.
## Step 3: Compare $ Q $ and $ K $
$ Q = 4 = K $, so at equilibrium.
# Answer: yes

### Second Equilibrium (Gray ⇌ Purple, $ K = \frac{1}{5} $)
# Explanation:
## Step 1: Count Moles
Gray: 8, Purple: 2.
## Step 2: Calculate $ Q $
$ Q = \frac{[\text{Purple}]}{[\text{Gray}]} = \frac{2}{8} = \frac{1}{4} $.
## Step 3: Compare $ Q $ and $ K $
$ Q = \frac{1}{4}
eq \frac{1}{5} = K $, so not at equilibrium.
# Answer: no

### Third Equilibrium (Light Blue ⇌ Purple, $ K = \frac{2}{3} $)
# Explanation:
## Step 1: Count Moles
Light Blue: 9, Purple: 1.
## Step 2: Calculate $ Q $
$ Q = \frac{[\text{Purple}]}{[\text{Light Blue}]} = \frac{1}{9} \approx 0.11 $. $ \frac{2}{3} \approx 0.67 $, $ Q
eq K $, so not at equilibrium. (Wait, correction: Wait, the reaction is Light Blue ⇌ Purple? Wait, the diagram: light blue (let's say A) and purple (B). Wait, maybe I misread. Wait, the third reaction: the left has light blue (A) and purple (B). Wait, count again: light blue (A) molecules: let's see, the light blue ones: 9? Wait no, the third box: light blue (A) and purple (B). Wait, the formula: $ K = \frac{[\text{B}]}{[\text{A}]} = \frac{2}{3} $. Wait, maybe I miscounted. Wait, the third grid: light blue (A) and purple (B). Let's count A (light blue): 9? No, looking at the third row left: the molecules: light blue (A) and purple (B). Wait, the purple one is 1, light blue is 9? Wait no, maybe the reaction is A ⇌ B, so $ K = \frac{[B]}{[A]} = \frac{2}{3} $. Wait, if B is 1, A is 9, $ Q = 1/9 \approx 0.11 $, $ K = 2/3 \approx 0.67 $, so $ Q
eq K $, so no. Wait, but maybe I miscounted. Wait, the third row left: the purple (B) is 1, light blue (A) is 9? Wait, no, let's count again. The third row left: the molecules: how many light blue (A) and purple (B). Let's see: the purple one is 1, light blue: let's count the light blue (A) molecules: each light blue is a cluster. Let's count: 9 light blue, 1 purple. So $ Q = 1/9 \approx 0.11 $, $ K = 2/3 \approx 0.67 $, so $ Q
eq K $, so not at equilibrium. So answer: no. Wait, but maybe the reaction is reversed? Wait, the reaction is A ⇌ B, $ K = 2/3 $. So $ K = [B]/[A] = 2/3 $. So if [B] = 1, [A] = 9, $ Q = 1/9
eq 2/3 $, so no.

### Fourth Equilibrium (Green ⇌ Beige, $ K = \frac{1}{3} $)
# Explanation:
## Step 1: Count Moles
Green (A): let's count the green molecules (the three - atom ones? Wait, no: the fourth reaction: green (A, three - atom) and beige (B, three - atom? Wait, no, the green is a three - atom (A) and beige is a three - atom (B)? Wait, the reaction is Green (A) ⇌ Beige (B), $ K = \frac{1}{3} $. Count Green (A): let's see, the green molecules: how many? The green clusters: let's count: 5? Wait no, the fourth box: green (A) and beige (B). Let's count: Green (A) molecules: 5? Beige (B) molecules: 3? Wait, no, let's look again. The fourth grid: green (A) (three - atom) and beige (B) (three - atom). Wait, the green ones: let's count: 5? Wait, no, the green clusters: each green is a three - atom. Let's count: green (A): 5? Beige (B): 3? Wait, no, the reaction is $ \text{Green}
ightleftharpoons \text{Beige} $, $ K = \frac{[\text{Beige}]}{[\text{Green}]} = \frac{1}{3} $. Let's count: Green (A) molecules: 5? Beige (B) molecules: 3? Wait, no, maybe I misread. Wait, the fourth box: green (A) and beige (B). Let's count again: Green (A) (three - atom clusters): let's see, the green ones: 5? Beige (B): 3? Wait, no, the green clusters: 5? Beige: 3? Then $ Q = \frac{3}{5} = 0.6 $, $ K = 1/3 \approx 0.33 $, so $ Q
eq K $, so not at equilibrium? Wait, no, maybe the reaction is reversed. Wait, $ K = \frac{1}{3} $, so if the reaction is $ \text{Green}
ightleftharpoons \text{Beige} $, $ K = \frac{[\text{Beige}]}{[\text{Green}]} = 1/3 $. So if [Beige] = 3, [Green] = 5, $ Q = 3/5 = 0.6
eq 1/3 \approx 0.33 $, so not at equilibrium. Wait, but maybe I miscounted. Wait, the beige (B) molecules: 3? Green (A) molecules: 5? So $ Q = 3/5 = 0.6 $, $ K = 1/3 \approx 0.33 $, so $ Q
eq K $, so no.

Wait, but let's re - do the third and fourth:

### Third Equilibrium (Correct Count)
Reaction: $ \text{A (light blue)}
ightleftharpoons \text{B (purple)} $, $ K=\frac{2}{3} $
Count A (light blue): 9, B (purple): 1
$ Q = \frac{[B]}{[A]}=\frac{1}{9}\approx0.11$, $ K = \frac{2}{3}\approx0.67$, $ Q
eq K $ → no

### Fourth Equilibrium (Correct Count)
Reaction: $ \text{A (green)}
ightleftharpoons \text{B (beige)} $, $ K = \frac{1}{3} $
Count A (green): 5, B (beige): 3
$ Q=\frac{[B]}{[A]}=\frac{3}{5} = 0.6$, $ K=\frac{1}{3}\approx0.33$, $ Q
eq K $ → no

But wait, maybe I made a mistake in the third reaction's molecule count. Let's re - check the third box: the left box (third row, first column) has light blue (A) and purple (B). The light blue molecules: let's count again. The light blue ones: each is a three - atom? Wait, no, the third box: the light blue molecules (A) are 9? Wait, no, looking at the image: the third row, first column: the molecules: light blue (A) and purple (B). The purple one is 1, light blue: let's count the light blue clusters: 9? Yes, so A = 9, B = 1. So $ Q = 1/9 $, $ K = 2/3 $, so no.

For the fourth row, first column (wait, no, fourth is second column). The fourth box (second column, third row): green (A) and beige (B). Green clusters: 5? Beige clusters: 3? So $ Q = 3/5 $, $ K = 1/3 $, so no.

But let's summarize the answers:

1. First reaction: yes
2. Second reaction: no
3. Third reaction: no
4. Fourth reaction: no

Wait, but maybe the third reaction's $ K $ is $ \frac{2}{3} $, and if A is 9 and B is 1, $ Q = 1/9
eq 2/3 $, so no.

Final Answers:

1. yes
2. no
3. no
4. no

(Note: The third and fourth might have counting errors, but based on the given $ K $ and molecule counts, this is the result.)

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QUESTION IMAGE

Question

Explanation:

First Equilibrium (Brown ⇌ Blue, \( K = 4 \))

Step 1: Count Moles

Step 2: Calculate Reaction Quotient (\( Q \))

Step 3: Compare \( Q \) and \( K \)

Step 1: Count Moles

Step 2: Calculate \( Q \)

Step 3: Compare \( Q \) and \( K \)

Step 1: Count Moles

Step 2: Calculate \( Q \)

Fourth Equilibrium (Green ⇌ Beige, \( K = \frac{1}{3} \))

Answer:

Second Equilibrium (Gray ⇌ Purple, \( K = \frac{1}{5} \))