For the hypothetical reaction, A2(g) + B2(g) ⇌ 2AB(g), ΔrG and ΔrS are 20 kJ/mole and –20 JK–1 mol–1 respectively at 200 K. If given for the reaction ΔrCp is 20 JK–1 mol–1 then ΔrH of the reaction at 400 K is :-
Question
For the hypothetical reaction, A2(g) + B2(g) ⇌ 2AB(g), ΔrG and ΔrS are 20 kJ/mole and –20 JK–1 mol–1 respectively at 200 K. If given for the reaction ΔrCp is 20 JK–1 mol–1 then ΔrH of the reaction at 400 K is :-
Solution
To solve this problem, we need to use the Gibbs-Helmholtz equation, which relates the change in Gibbs free energy (ΔrG), the change in enthalpy (ΔrH), and the change in entropy (ΔrS) at a given temperature (T). The equation is:
ΔrG = ΔrH - TΔrS
We know that at 200K, ΔrG = 20 kJ/mol and ΔrS = -20 J/K*mol. We can substitute these values into the equation to solve for ΔrH at 200K:
20 kJ/mol = ΔrH - (200K)(-20 J/K*mol) 20 kJ/mol = ΔrH + 4 kJ/mol ΔrH = 20 kJ/mol - 4 kJ/mol = 16 kJ/mol
Now, we need to find ΔrH at 400K. We know that ΔrCp = 20 J/K*mol, which is the change in heat capacity at constant pressure. The change in enthalpy with temperature can be calculated using the equation:
ΔrH(T2) = ΔrH(T1) + ∫(T1 to T2) ΔrCp dT
Assuming ΔrCp is constant over the temperature range, the equation simplifies to:
ΔrH(400K) = ΔrH(200K) + ΔrCp*(T2 - T1) ΔrH(400K) = 16 kJ/mol + (20 J/Kmol)(400K - 200K) ΔrH(400K) = 16 kJ/mol + 4 kJ/mol = 20 kJ/mol
So, the change in enthalpy of the reaction at 400K is 20 kJ/mol.
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