Spiker

10+ Year Member
Feb 13, 2009
610
3
Status
Pre-Medical
So I run into a problem on a test and I am pretty sure i am right

A+B->C+D Fast

D+B->E Slow

K1[A]=K-1[C][D]
[D]=K1/K-1[A]/[C]
K2[D]=E
K2[ K1/K-1[A]/[C]]=E
K [A]/[C]=E
K^2[A]/[C]=E<---my answer
K^2[A]=E<---Test answer
 

bruceleehiiiyaa

Membership Revoked
Removed
10+ Year Member
Oct 6, 2008
294
0
Status
So I run into a problem on a test and I am pretty sure i am right

A+B->C+D Fast

D+B->E Slow

K1[A]=K-1[C][D]
[D]=K1/K-1[A]/[C]
K2[D]=E
K2[ K1/K-1[A]/[C]]=E
K [A]/[C]=E
K^2[A]/[C]=E<---my answer
K^2[A]=E<---Test answer



is this kaplan? i had a similar prob too. i dont think those bastards took the reverse K-1 rxn into account, which they should have. if they didnt, that explains how they arrived to their conclusion.
 
OP
S

Spiker

10+ Year Member
Feb 13, 2009
610
3
Status
Pre-Medical
is this kaplan? i had a similar prob too. i dont think those bastards took the reverse K-1 rxn into account, which they should have. if they didnt, that explains how they arrived to their conclusion.
LOL yea!! you got good memory
 

ezsanche

10+ Year Member
5+ Year Member
Aug 15, 2007
594
0
Status
So I run into a problem on a test and I am pretty sure i am right

A+B->C+D Fast

D+B->E Slow

K1[A]=K-1[C][D]
[D]=K1/K-1[A]/[C]
K2[D]=E
K2[ K1/K-1[A]/[C]]=E
K [A]/[C]=E
K^2[A]/[C]=E<---my answer
K^2[A]=E<---Test answer


i got the same as you did