Indy chem ~ eq. const. (1 Viewer)

hellohello

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this is how u did it

initially 0.25 + 0.12 = 0

at equil (0.25-0.05) + (0.12-0.025) = 0.05

there fore this equals

[0.05]^2/
[(0.20)^2 x 0.095]

= 0.63
 

vegeta_316

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800

I don't have the paper with me so I forget the concentrations of the reactants and products initially and at equilibrium but I drew that table:

initial
change
equilibrium

and filled in the required details of the products and reactants.

Pretty sure it came out to 800 which looks like a good k value to me.
 
Last edited:

Nuggy

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K was 800

that was wat almost all the guys in my class got... (sbhs) so... i'm feeling pretty confident about that answer =D
 

Nuggy

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was the last qusetion fulla crap or is it just my impression...

i picked... solvay process and contact process . . .
 

vegeta_316

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Originally posted by Nuggy
was the last qusetion fulla crap or is it just my impression...

i picked... solvay process and contact process . . .
Yeah I chose the same processes for the base and acid respectively. There was plenty to talk about environmental wise, with the dispsoal of things like SO2 an acidic oxide into the atmosphere, CaCl2 waste in the solvay process etc etc.
 

Misturi

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l didn't do it. I found the entire test hard. I only did a) i) in industrial chemistry. :(
 

forsaken_99

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I got 40.

Originally i got 800, but then i re-did it using the mol of NO and comparing it with the ratio.
 

Misturi

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Originally posted by forsaken_99
I got 40.

Originally i got 800, but then i re-did it using the mol of NO and comparing it with the ratio.
How much would you be kicking yourself if the answer is 800? But then again, how much would you be cheering if the answer was 40? :)
 

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