Find the temperature at which the reaction below is spontaneous given this data:
Hf^o of NH3(g) = - 46 kJ/mol, S^o of NH3(g) = 192.5 J/mol·K, S^o of N2(g) = 191.5 J/mol·K,
S^o of H2(g) = 130.6 J/mol·K.
N2(g) + 3H2(g) <-----> 2NH3(g)
delta G = delta H - TdeltaS
delta H you have.
delta Srxn = Sproducts - Sreactants
Solve for T which will give a negative G.
To find the temperature at which the reaction is spontaneous, we can use the Gibbs free energy equation:
ΔG = ΔH - TΔS
Where:
ΔG is the Gibbs free energy change,
ΔH is the enthalpy change,
T is the temperature in Kelvin,
ΔS is the entropy change.
In this case, we need to consider the entropy change for the reaction, which is given by:
ΔS = ΣnS^o(products) - ΣmS^o(reactants)
Where:
n and m are the coefficients in the balanced chemical equation,
S^o(products) and S^o(reactants) are the standard entropies of the products and reactants, respectively.
Let's calculate ΔS for the reaction:
n = 2 (for NH3)
m = 1 (for N2) + 3 (for H2)
ΔS = (2 * S^o of NH3) - (1 * S^o of N2) - (3 * S^o of H2)
= (2 * 192.5 J/mol·K) - (1 * 191.5 J/mol·K) - (3 * 130.6 J/mol·K)
Now, we can substitute the given values into the equation and solve for ΔG:
ΔG = -46 kJ/mol - TΔS
Since we want to find the temperature at which the reaction is spontaneous, ΔG should be less than zero (ΔG < 0). So we can set up the inequality:
ΔG < 0
Now, substitute the given value of ΔG and solve for T:
-46 kJ/mol - TΔS < 0
Solve for T:
T > (-46 kJ/mol) / (ΔS/mol)
Substitute the calculated value of ΔS to find the temperature at which the reaction is spontaneous.