Question

Start from the following expression that relates the equilibrium constant K to the standard free energy (on a molecular basis)

We know that , where ∆H⁰ and ∆S⁰ are the standard enthalpy and entropy, respectively. The standard enthalpy and standard entropy are constants.

From the above information, find an expression for the slope of ln(K) when plotted against (1/T). In other words, find an expression for

Answer 1

Solution :

the following expression that relates the equilibrium constant K to the standard free energy :

∆Go = -RT lnK...........................1

where ∆Go = standard free energy

We know that ∆Go = ∆Ho  - T ∆So .........................2

where ∆H0 and ∆S0 are the standard enthalpy and entropy.

From equation 1 and 2,

∆Ho  - T ∆So = -RT lnK

lnK = -(∆Ho  - T ∆So) / (RT)

lnK = -∆Ho/ RT  + ∆So/R

Com pare this equation with y = mx + c,

y = lnK,

x = 1/T,

Now if we draw a line of y vs x we get,

Slope of line = m = -∆Ho/ R

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