What Is the Equilibrium Constant Kc for This Reaction?

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Discussion Overview

The discussion revolves around calculating the equilibrium constant (Kc) for the reaction 2NS(g) + 2H2(g) <--> N2(g) + 2H2S(g) under specific conditions. Participants are addressing a homework problem involving initial and equilibrium concentrations in a 4 L flask at 25°C.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents the initial conditions and attempts to calculate Kc, initially suggesting Kc = 1 based on the equilibrium concentrations.
  • Another participant questions the correctness of the reaction quotient and suggests verifying the initial amounts and changes in concentrations.
  • A later reply points out a potential typo in the chemical formula and emphasizes that if the amount of NS has not changed, the other reactants' amounts would also remain unchanged.
  • One participant confirms the initial concentration calculation as 0.625 M based on the initial moles divided by the volume of the flask.
  • There is a correction regarding the chemical species involved, with one participant acknowledging a typo and reiterating the problem statement.
  • Ultimately, one participant revises their answer to Kc = 1.6, indicating a change in their calculation.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the value of Kc, with differing calculations presented. Uncertainty remains regarding the interpretation of the equilibrium conditions and the correct application of the reaction quotient.

Contextual Notes

There are unresolved issues regarding the initial and equilibrium concentrations, as well as potential typos in the chemical formulas. The discussion reflects varying interpretations of the problem setup.

phys1618
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Homework Statement


2NS(g) + 2H2(g)<---> N2(g) + 2H2S(g) @ equilibrium in a 4.L flask @ 25o
C. 2.5 mol of eah component initially is put into the flask. At equilibrium, 2.5 mol of NS is present. what is Kc for the rxn at this temp??

Homework Equations



Kc=[products]/[reactants]
M=moles/L
initial - change = equilibrium

The Attempt at a Solution



Kc=1 i thought the answer is a little odd... because the intital and equilbrium is 2.5 moles for NS so I'm unsure about my answer... please help! geatly appreciates all help! thank you in advance.
if my answer is wrong...can someone show me how to do it i know how to set up Kc, but I am troubled when finding the intial, change and equilibrium of each components.
 
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First of all - can you write correct reaction quotient?
 
kc=[N2][HS]2/ [NS]2

2

 
OK (let's assume HS is just a typo :wink:).

Question is a little bit strange, as it seems amount of NS have not changed. But that's not your problem - if amount of NS have not changed, there were no reaction, thus initial amount of other reactants have not changed either. Can you calculate what are their concentrations?

PS PLease check - just in case - if you have copied question exactly as it was written.
 
hehe yea i did make a typo..sorry
their initial concentration is .625 M. I took 2.5 n divided that by 4.0 L
2NS9g0 +2H2(g<-----> N2g) + 2H2S(g) at equilibrium in a 4.0L flask at 25oC. suppose 2.5 mol of each component initially is put into the flask. At equilibrium, 2.5 mol of NO is present. What is Kc for the reaction at this temperature?
 
phys1618 said:
2NS(g) + 2H2(g)<---> N2(g) + 2H2S(g) @ equilibrium in a 4.L flask @ 25o
C. 2.5 mol of eah component initially is put into the flask. At equilibrium, 2.5 mol of NS is present. what is Kc for the rxn at this temp??

phys1618 said:
2NS9g0 +2H2(g<-----> N2g) + 2H2S(g) at equilibrium in a 4.0L flask at 25oC. suppose 2.5 mol of each component initially is put into the flask. At equilibrium, 2.5 mol of NO is present. What is Kc for the reaction at this temperature?

Sigh...

Put know concentrations into reaction quotient.
 
kc is still 1
 
nvm its 1.6
 

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