Answer:
- 1.124 x 10⁵ J = -1.124 x 10² kJ
Step-by-step explanation:
The change in Gibbs free energy is related to the equilibrium constant K by
ΔGº = - RTlnK
But, we also know that by definition
ΔGº = ΔHº - TΔSº
thus
ΔHº - TΔSº = - RTlnK
lnK = - ΔHº/RT + ΔSº/R
This equation has the form y = mx + b where m is the slope and b is the y intercept.
Therefore, a plot lnK versus 1/T will give us a straight line with slope , m= - ΔHº/R.
We know m = 1.352 x 10⁴ K
so we can solve for ΔHº
ΔHº = - mR = 1.352 x 10⁴ K x 8.314 J / K = - 1.124 x 10⁵ J = -1.124 x 10² kJ
The standard enthalpy of the reaction is -112.4 kJ/mol.
The equation of a straight line is given by, y = mx + c
m = slope
c = y-intercept
Consider the equation;
lnK = -ΔH°/RT + ΔS°/R
Where;
K = equilibrium constant
R = gas constant
T = Absolute temperature
ΔH° = Standard enthalpy of reaction
ΔS° = Standard entropy of reaction
Comparing the two equations;
-ΔH°/RT = mx
So;
-ΔH°/R = 1.352 x 10^4 K
Since R = 8.314 JK-1mol-1
ΔH° = -(1.352 x 10^4 K × 8.314 JK-1mol-1)
ΔH° = -112.4 kJ/mol
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