At what temperature will the following reaction happen spontaneously, given that ΔH=8.91×103J and ΔS= –219.20 J/K?

CaSO4(s) + 2HCl(g) → CaCl2(s) + H2SO4(l)

Calculate dH for the reaction and dS for the reaction.

Then dGrxn = dHrxn - TdSrxn
Set dGrxn = 0, substitute the values you have for dHrxn and dSrxn and solve for T.

not spontaneous

To determine at what temperature the reaction will happen spontaneously, we need to calculate the Gibbs free energy change (∆G) using the equation:

∆G = ∆H - T∆S

Where:
∆G is the Gibbs free energy change
∆H is the enthalpy change
T is the temperature in Kelvin
∆S is the entropy change

In this case, we are given ∆H = 8.91×10^3 J and ∆S = -219.20 J/K. We need to find the temperature at which ∆G is zero (∆G = 0), which is the temperature at which the reaction becomes spontaneous.

Setting ∆G = 0 and rearranging the equation, we get:

0 = ∆H - T∆S

Rearranging further to isolate T, we have:

T∆S = ∆H

T = ∆H/∆S

Now we can substitute the values:

T = (8.91×10^3 J)/(-219.20 J/K)

Calculating this equation will give us the temperature at which the reaction happens spontaneously.