How Do You Determine the Equilibrium Constant for Modified Reactions?

Click For Summary

Discussion Overview

The discussion revolves around determining the equilibrium constant for modified chemical reactions, specifically focusing on how to derive the equilibrium constants for given reactions based on known values. The scope includes mathematical reasoning and conceptual clarification related to chemical equilibria.

Discussion Character

  • Mathematical reasoning
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant presents a problem involving the equilibrium constant for the reaction H2 + I2 → 2 HI, asking for the value of Keq for the modified reaction 1/2 H2 + 1/2 I2 → HI.
  • Another participant suggests that the second reaction is half of the first, leading to a derived equilibrium constant of x1/2, where x is the original Kc.
  • A participant mentions that writing the reaction quotient for each reaction is a straightforward approach to understanding the equilibrium constants.
  • One participant states that the equilibrium constant for the second reaction should be 1/k2, questioning why it wouldn't be 1/k or -1/k2 instead.
  • Another participant challenges the reasoning behind the proposed values, suggesting that comparing reaction quotients should clarify the new equilibrium constant's value.
  • A later reply acknowledges the reasoning about the inverse reaction and the doubling effect, affirming the 1/k2 conclusion.
  • One participant offers assistance with notation for equilibrium reactions, providing symbols for clarity.

Areas of Agreement / Disagreement

Participants express differing views on the derivation of equilibrium constants for modified reactions, with no consensus reached on the reasoning behind the proposed values.

Contextual Notes

Some participants express uncertainty regarding the mathematical relationships between the equilibrium constants and the modifications of the reactions, indicating a need for further clarification on the underlying principles.

shaa
Messages
6
Reaction score
0
Hello, I'm new to the forums and did in fact search for my thread before posting. I couldn't find anything that helped me with what I'm looking for. With that said, the problem I'm having is:

1) The value of Keq for the equilibrium H2+I2 → 2 HI is 794 at 25°C. What is the value of Keq for the equilibrium below?
1/2 H2+1/2 I2 → HI

2) The equilibrium constant for reaction 1 is K. The equilibrium constant for reaction 2 is _____.
1: SO2+1/2 O2 → SO3
2: 2SO3 → 2SO2+O2

I apologize for not denoting an equation at equilibrium, but I couldn't find the right arrow in the "Quick Symbols" box. Both equations are at equilibrium and involve only matter in the gaseous state.
 
Physics news on Phys.org
I've gotten the first question, considering that the second reaction is half of the first. Resulting in: x1/2, and x being the original Kc.

I know what the answer to the first question is, but I'm having difficulty understanding why it is what it is.
 
Last edited:
This is a simple math. Write reaction quotient for each reaction.
 
It states that the equilibrium constant for reaction 1 is = k. The equilibrium constant for reaction 2 is = 1/k2.

The answer makes complete sense given a general form of an equilibrium reaction, but why wouldn't the first reaction be = 1/k or the second be = -1/k2?
 
I don't understand why you think it should be. If you write reaction quotients for each reaction and compare them, it is obvious what value the new equilibrium constant has.

You can as well ask why \sqrt 2 doesn't equal \pi.
 
Hahahah. Of course. It's the inverse reaction and it double. Thus 1/k2.
Thanks? :P
 
shaa said:
I apologize for not denoting an equation at equilibrium, but I couldn't find the right arrow in the "Quick Symbols" box. Both equations are at equilibrium and involve only matter in the gaseous state.
One of these might help; feel free to copy-and-paste for future use, or bookmark the links in my sig:
← or ↔

p.s. Welcome to Physics Forums.
 
Last edited:

Similar threads

Replies
4
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
Replies
15
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
5
Views
2K
  • · Replies 79 ·
3
Replies
79
Views
9K
Replies
2
Views
4K
Replies
3
Views
2K