for the reaction at 298 K 2A+b ---->C
enthalpy=100 K/cal entropy= 0.050 kcal/k
assuming enthalpy and entropy to be constant over the temperature range, at what temperature will the reaction become spontaneous?
answer: T>2000K
Thomas, it wouldn't hurt to read over your posts before you push that submit button. No caps and no periods and typos to boot leave a post that's difficult to interpret. For this one I've guessed at half of the question; assuming I've guessed right you do the problem this way.
dG = dH - TdS
At the spontaneous stage, dG must be negative so substitute zero, the tipping point, for dG.
That gives you TdS = dH. Substitute the values for dS and dH and solve for T.
thanks
To determine at what temperature the reaction will become spontaneous, we can use the Gibbs Free Energy equation:
ΔG = ΔH - TΔS
Where ΔG is the change in Gibbs Free Energy, ΔH is the change in enthalpy, T is the temperature in Kelvin, and ΔS is the change in entropy.
For the reaction to become spontaneous, ΔG must be negative. Let's plug in the given values:
ΔH = 100 Kcal
ΔS = 0.050 Kcal/K
ΔG = 100 - T * 0.050
For the reaction to become spontaneous, ΔG < 0, so:
100 - T * 0.050 < 0
Simplifying the equation:
100 < T * 0.050
Dividing both sides of the inequality by 0.050:
2000 < T
Therefore, the reaction will become spontaneous at a temperature greater than 2000 K.