3600 K
-219.95
For the reaction to be spontaneous ∆G needs to be <0∆G = ∆H - T∆S
∆G = 0 = -220 - T(-0.05)
0.05T = 220
T = 4400ºK
So, temperature would have to be less than 4400ºK
It is not spontaneous.
it can never be spontanious
10,267 kJ are needed
-127 KJmol-1
I think you should try and use q = mcT where q = heat change m = mass of substance c = specific heat capacity of the substance ( Water is 4.187 kJmol-1) T = change in temperature (oC) so for your question, q = mcT q = 12 x 4.187 x (93-15.4) q = 3898.9344 kJmol-1 Hope this helps!!
I2(s) --> I2(g); dH=62.4kJ/mol; dS=0.145kJ/mol. The reaction will favor the product at this temperature. Your entropy is positive and your enthalpy is also positive, so this reaction will not be spontaneous at all temperatures. But at room temperature, which is 298K, it will be spontaneous and proceed left to right. (this is the sublimation of iodine)
400 K
It is not spontaneous.
it is never spontaneous
it can never be spontanious
G = 0 kJ/mol
It is spontaneous
G= 0 kJ/mol
It can be said to be exothermic since H is negative. Also below a temp of 3708ºC, the reaction will be spontaneous because ∆G will be negative. This is from ∆G = ∆H - T∆S.
it can never be spontanious
H2(g) + S(s) —> H2S + 20.6 kJ
just add them together and you get 147kj