Partial pressures and equilibrium question

AI Thread Summary
The discussion revolves around calculating the equilibrium partial pressures for the reaction 2HI (g) ⇌ H2 (g) + I2 (g) with Kp = 0.02 at 445 degrees C. The initial partial pressures are given as 0.05, 0.01, and 0.01 atm. The calculated reaction quotient Qp is found to be greater than Kp, indicating the reaction shifts to the left. Attempts to solve the quadratic equation for equilibrium pressures yield inconsistent results, leading to confusion and requests for assistance. The main issue highlighted is the need for careful mathematical verification to arrive at the correct equilibrium values of 0.0546, 0.0077, and 0.0077 atm.
veitch
Messages
7
Reaction score
0

Homework Statement



For the reaction 2HI (g) <-- --> H2 (g) + I2 (g) Kp = .02 at 445 degrees C
Parital pressures are .05, .01 and .01 respectively. Perform a calc to see which way it goes then figure out the equilibrium partial pressures of the gasses

The Attempt at a Solution



Qp = (.01)(.01)/(.05^2) = .0001/.0025 = .04

Since Qp>Kp reaction goes left

I .05 .01 .01
C +2x -x -x
E .05+2x .01-x .01-x

.02 = (.01-x)^2/(.05+2x)
0 = -x^2 + .06x + .0009

Quadratic equation yields -.0264 and -.0336... neither of which make sense as the final answers are given and they should be:
.0546, .0077 and .0077 atm
Which makes x = .0023...

That was my third time running through this problem and each time I got a different answer... no idea how I screwed this one up. it's only the first problem on an assignment due tomorrow... any help would be greatly appreciated.
 
Physics news on Phys.org
veitch said:
.02 = (.01-x)^2/(.05+2x)

This is OK.

Quadratic equation yields -.0264 and -.0336

This is not. Just check your math.
 
Borek said:
This is not. Just check your math.

You're right.. it should be 0= x^2 -.06x -.0009
Which has roots of 0.072 and -0.012 ...

But .05 + 2(.072) = 0.194 yet it should equal .0546 atm :(
 
while (x != 0.0023) check_your_math;
 
Thread 'Confusion regarding a chemical kinetics problem'
TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...
I don't get how to argue it. i can prove: evolution is the ability to adapt, whether it's progression or regression from some point of view, so if evolution is not constant then animal generations couldn`t stay alive for a big amount of time because when climate is changing this generations die. but they dont. so evolution is constant. but its not an argument, right? how to fing arguments when i only prove it.. analytically, i guess it called that (this is indirectly related to biology, im...
Back
Top