Please help!!
Predict whether N2(g) + 3H2(g) ⇌ 2NH3(g) is spontaneous at 773.15 K.
Assume: ∆H = -92.22 kJ and ∆S = -198.75 J/mol·K. Describe all components.
Show work please!!
dG = dH - TdS
The problem gives you dH and dS and T. Solve for dG. If dG < 0 the reaction is spontaneous. If dG > 0 it is not spontaneous.
@DrBob222 Thank you Sir this helped a lot
To determine whether the reaction N2(g) + 3H2(g) ⇌ 2NH3(g) is spontaneous at 773.15 K, we can use the Gibbs Free Energy equation:
ΔG = ΔH - TΔS
Where:
ΔG = Gibbs Free Energy
ΔH = Enthalpy change
T = Temperature
ΔS = Entropy change
Given:
ΔH = -92.22 kJ (convert to J by multiplying by 1000: ΔH = -92.22 * 1000 J = -92220 J)
ΔS = -198.75 J/mol·K
T = 773.15 K
Plugging in the values into the equation, we get:
ΔG = -92220 J - (773.15 K)(-198.75 J/mol·K)
Now we can solve for ΔG.