fiitjee ft XI...plz sum1

does delta G(not) change with temp...thou it is defined at 298K only

.Q.. 2 A(g) + B(g) (equlibrium) 3S(s) + T(g)
delta H(not) = -145

At 298K forward rxn is spontaneous, wt change if ny wud occur in deltaG(not) value as temp. is inc
.a. becomes positive
b. doesnt change
c. becomes neg.

13 Answers

1
greatvishal swami ·

Q2 b) no change

1
greatvishal swami ·

delta G0 becomes +ve and -ve acc to temperature for ex in an exothermic reaction on inc temp reaction may become non spontaneous and δG0 might become +ve

13
deepanshu001 agarwal ·

yur ans and explanation dont corelate

1
vector ·

c

13
deepanshu001 agarwal ·

ppl plz giv explanations to support ur ans.....

1
greatvishal swami ·

deepanshu in that Q of urs i dont think if spontanity is playing any role thats y any change in temp will not effecy G0

13
deepanshu001 agarwal ·

u mean 2 say delta G(not) doesnt change wid temp....

dats wt i tryin 2 gt a clr idea abt

13
deepanshu001 agarwal ·

the ans is dat it becomz more positive can ny explain...

i m nt gettin it at ol

13
deepanshu001 agarwal ·

cant use that relation... bcz even K changes wid temp
using
dG(not)=d H(not) - dS(not)
u do get the ans given bt wt i intend to ask is that is that dG(net) is changin
wid change in temp.. though it is defined only at 298K....????

1
vector ·

ACTUALLY as h is -ve so reaction is exo an on increasingt it ll go backward n thus G° going +

1
vector ·

sry deleted tat post

13
deepanshu001 agarwal ·

i already noe that but.....
how is it so dG(not) is changin wid temp....???

1
vector ·

as temp is increasing n rate of backward n forward reactin is changing so suppse at a higher temp t" a new value of G° will be defined giving idea abt fissiblity of f reaction at that temp

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