38. If the equilibium constants for the two reactions at 400 K: + ICl3 (g) = ICl;Br2 (g) Br2 (g) 2 ICl;Br2 (g) + Br2 (g) = 2 IB13 (g) +3 Cl2 (g) 2.5 x 10 4.0 x 10-7 K2 then what is the equilibrium constant for the following reaction ? = 2 IB13 (g) + 3 Cl2 (g) 3 Br2 (g) + 2 ICI3 (g) K= ?
38. If the equilibium constants for the two reactions at 400 K: + ICl3 (g) = ICl;Br2 (g) Br2 (g) 2 ICl;Br2 (g) + Br2 (g) = 2 IB13 (g) +3 Cl2 (g) 2.5 x 10 4.0 x 10-7 K2 then what is the equilibrium constant for the following reaction ? = 2 IB13 (g) + 3 Cl2 (g) 3 Br2 (g) + 2 ICI3 (g) K= ?
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Chapter1: Chemical Foundations
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38) please see attached
![**Equilibrium Constants in Chemical Reactions**
**Problem Statement:**
Given the equilibrium constants for two reactions at 400 K:
1. \( \text{Br}_2 (g) + \text{ICl}_3 (g) \rightleftharpoons \text{ICl}_3\text{Br}_2 (g) \) with an equilibrium constant \( K_1 = 2.5 \times 10^3 \)
2. \( 2 \text{ICl}_3 \text{Br}_2 (g) + \text{Br}_2 (g) \rightleftharpoons 2 \text{IBr}_3 (g) + 3 \text{Cl}_2 (g) \) with an equilibrium constant \( K_2 = 4.0 \times 10^{-7} \)
Determine the equilibrium constant for the following reaction:
\[ 3 \text{Br}_2 (g) + 2 \text{ICl}_3 (g) \rightleftharpoons 2 \text{IBr}_3 (g) + 3 \text{Cl}_2 (g) \]
**Solution:**
To find the equilibrium constant (\( K \)) for the overall reaction, we need to combine the given equilibrium reactions.
First, we multiply the first reaction by 2:
\[ 2 \text{Br}_2 (g) + 2 \text{ICl}_3 (g) \rightleftharpoons 2 \text{ICl}_3\text{Br}_2 (g) \]
The new equilibrium constant for this reaction will be \( (K_1)^2 = (2.5 \times 10^3)^2 = 6.25 \times 10^6 \)
Next, we sum this transformed reaction with the second given reaction:
\[ 2 \text{ICl}_3 \text{Br}_2 (g) + \text{Br}_2 (g) \rightleftharpoons 2 \text{IBr}_3 (g) + 3 \text{Cl}_2 (g) \]
Which simplifies to:
\[ 3 \text{Br}_2 (g) + 2 \text{ICl}_3 (g) \rightleftharpoons 2 \text{IBr}_3 (g) +](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F48deb947-3d41-43de-ac43-bef03e01cc59%2F4091cb4f-6d3e-4d0c-be3d-c01e86ed39f2%2F9tquaxb.png&w=3840&q=75)
Transcribed Image Text:**Equilibrium Constants in Chemical Reactions**
**Problem Statement:**
Given the equilibrium constants for two reactions at 400 K:
1. \( \text{Br}_2 (g) + \text{ICl}_3 (g) \rightleftharpoons \text{ICl}_3\text{Br}_2 (g) \) with an equilibrium constant \( K_1 = 2.5 \times 10^3 \)
2. \( 2 \text{ICl}_3 \text{Br}_2 (g) + \text{Br}_2 (g) \rightleftharpoons 2 \text{IBr}_3 (g) + 3 \text{Cl}_2 (g) \) with an equilibrium constant \( K_2 = 4.0 \times 10^{-7} \)
Determine the equilibrium constant for the following reaction:
\[ 3 \text{Br}_2 (g) + 2 \text{ICl}_3 (g) \rightleftharpoons 2 \text{IBr}_3 (g) + 3 \text{Cl}_2 (g) \]
**Solution:**
To find the equilibrium constant (\( K \)) for the overall reaction, we need to combine the given equilibrium reactions.
First, we multiply the first reaction by 2:
\[ 2 \text{Br}_2 (g) + 2 \text{ICl}_3 (g) \rightleftharpoons 2 \text{ICl}_3\text{Br}_2 (g) \]
The new equilibrium constant for this reaction will be \( (K_1)^2 = (2.5 \times 10^3)^2 = 6.25 \times 10^6 \)
Next, we sum this transformed reaction with the second given reaction:
\[ 2 \text{ICl}_3 \text{Br}_2 (g) + \text{Br}_2 (g) \rightleftharpoons 2 \text{IBr}_3 (g) + 3 \text{Cl}_2 (g) \]
Which simplifies to:
\[ 3 \text{Br}_2 (g) + 2 \text{ICl}_3 (g) \rightleftharpoons 2 \text{IBr}_3 (g) +
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