For a reversible reaction taking place at 127 degree celsius and 1ATM pressure.delta G=23.5kj/mol. for the forward reaction,calculate value of eqm.constant for the reaction
\[\LARGE \delta G=-2.303RTlogK\] I used this,getting wrong answer.
\[\LARGE 23500=-2.303 \times 8.314 \times 400 \times logK\]
relation to yahi hota hai mere khayaal.
\[\LARGE K=10^{0.30}\] gettting something likethis
not sure if i substituted correctly,the values
127 degree Celsius hai. convert it into kelvin
acha kiya hai. thik hai.
did,400K
e^7.07 hai answer?
arre numerically nikaalo
e wagera nahi :/
1176.14
its 8.59*10^-4
dekhta hun. get back to you later.
dont do it after 12AM :P
aa to rha hai. kya problem hai? calculator se krna pdega.
kya aya logk?
and meri substituted values sahi thi? yuhuuuuuuuuuuu @yrelhan4
haan haan thik thi. mereko yaad nhi kya aaya tha. pehle log(1/K) niklega. kyun ki -logK=log(1/K), fir 10^(---)=abc bhi krke 1/K niklega, fir 1/abc krke K niklega.
okay :|
TY
yw. :/
it'll be -23500 = -RT lnK
why -23500?
forward reacn would imply dG < 0
its a reversible reaction.
question wants us to calc K for forward reacn
and delta G for the forward reaction is given. if it was negative it would have been given
dG for forward reacn ? o.O It means nothing. dG for the whole reacn is given. for it to proceed forward, dG< 0 .
hmm. :O but i still have my doubts. when you say deltaG=-nFE(electrochem), and you are given a deltaG for a reaction, and you are a deltaG, do you do -deltaG then too? i think the negative sign on the RHS takes care of that. i dont know. i may be wrong.
Well anytime dG is put with the sign. The sign is always given in the ques. This was a different ques which asks us to calc K for forward reacn, Well I don't understand the ques practically, but theoretically, you will have to take dG<0. I have cross checked the ans via that.
-23500 = -RT lnK gives you 8.59*10^-4?
dafaq! it doesn't! o.O
I was wrong1 -_-
got carried away, so sorry! :|
haha.. np. :P
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