Equilibrium Chemistry Homework: Find [HI] at 400°C

Click For Summary
SUMMARY

At 400°C, the equilibrium constant Kc for the reaction H2 + I2 = 2HI is 64. When 3.0 mol of H2 and 3.0 mol of I2 are placed in a 4.0 L flask, the initial molarity of both reactants is 0.75 M. To find the equilibrium concentration of HI, the user must account for the stoichiometry of the reaction, which indicates that the concentrations of H2 and I2 will decrease as HI is produced. The correct equilibrium concentration of HI is 1.2 M, as confirmed by proper application of the equilibrium expression.

PREREQUISITES
  • Understanding of chemical equilibrium concepts
  • Knowledge of stoichiometry and molarity calculations
  • Familiarity with equilibrium constant expressions
  • Basic algebra skills for solving equations
NEXT STEPS
  • Review the concept of equilibrium constants in chemical reactions
  • Practice stoichiometric calculations in equilibrium scenarios
  • Learn how to derive and manipulate equilibrium expressions
  • Explore the effects of concentration changes on equilibrium position
USEFUL FOR

Chemistry students, educators, and anyone studying chemical equilibrium and reaction dynamics will benefit from this discussion.

ColinTI89
Messages
2
Reaction score
0

Homework Statement


At 400 degrees celsius, Kc=64 for the equilibrium H2 + I2 = 2HI. If 3.0 mol H2 and 3.0 mol of I2 are introduced into an empty 4.0 L flask, find the equilibrium concentration of HI at 400 degrees celsius.


Homework Equations


[HI]^2
--------

[I2]


The Attempt at a Solution



The answer given in the answer key is 1.2 M, which I am not getting. Since molarity is defined as mol/Liter, I calculate the molarity of the two given substances which is both 0.75 M. I set up the equation like this (sorry for the crappy quality):

...[HI]^2
64 = ----------
...[0.75] [0.75]

Multiplying 0.75 x 0.75 gives 0.5625 on the bottom. I then multiply both sides by 0.5625 to isolate [HI]^2. It now looks like this:

36=[HI]^2.

When I square root both sides I get 6.0, which doesn't sound correct anyways. Can anyone help? P.S. It's my first post on PhysicsForums!
 
Last edited:
Physics news on Phys.org
Keep in mind that the same reaction that produces HI also uses up both I2 and H2. Those elements may initially have concentrations of 0.75 mol/L, but this will decrease as they are consumed by the reaction.

Try writing a balanced equation for the reaction, and use the stoichiometric ratios to determine the amount of each element left after you've produced x moles of HI.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
15
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
6
Views
14K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 8 ·
Replies
8
Views
4K