1. Write the decomposition reaction

$2HI(g) \rightleftharpoons H_2(g)+I_2(g)$

We are told that at equilibrium, 40% of HI decomposes at $300\,K$.

2. Take initial moles and find equilibrium moles

Assume initially:

$HI=1\text{ mol}, \qquad H_2=0, \qquad I_2=0$

If $40\%$ of $HI$ decomposes, then moles of $HI$ consumed are:

$0.4$

From

$2HI \to H_2+I_2$

if $0.4$ mol of $HI$ is consumed, then extent gives:

$H_2 = I_2 = \frac{0.4}{2}=0.2$

So equilibrium moles are:

$HI=0.6, \qquad H_2=0.2, \qquad I_2=0.2$

Total moles at equilibrium:

$0.6+0.2+0.2=1.0$

Hence mole fractions are the same as the moles:

$y_{HI}=0.6, \quad y_{H_2}=0.2, \quad y_{I_2}=0.2$

Since total pressure is $1\,\text{atm}$,

$p_{HI}=0.6\,\text{atm}, \quad p_{H_2}=0.2\,\text{atm}, \quad p_{I_2}=0.2\,\text{atm}$

3. Find equilibrium constant

For the reaction

$2HI(g) \rightleftharpoons H_2(g)+I_2(g)$

$K_p=\frac{p_{H_2}\,p_{I_2}}{(p_{HI})^2}$

Substitute values:

$K_p=\frac{(0.2)(0.2)}{(0.6)^2}$

$K_p=\frac{0.04}{0.36}=\frac{1}{9}$

Because $\Delta n = (1+1)-2=0$, we have:

$K_p=K$

So,

$K=\frac{1}{9}$

4. Use relation between standard Gibbs free energy and equilibrium constant

$\Delta G^\circ = -RT\ln K$

Thus,

$\Delta G^\circ = - (8.31)(300)\ln\left(\frac{1}{9}\right)$

Since

$\ln\left(\frac{1}{9}\right)=-\ln 9$

so

$\Delta G^\circ = (8.31)(300)\ln 9$

Now,

$\ln 9 = 2\ln 3$

Given:

$\log 3=0.477, \qquad \ln 10 = 2.3$

Therefore,

$\ln 3 = 2.3\times 0.477 = 1.0971$

Hence,

$\ln 9 = 2(1.0971)=2.1942$

Now calculate:

$\Delta G^\circ = 8.31\times 300\times 2.1942$

$8.31\times 300 = 2493$

$\Delta G^\circ = 2493 \times 2.1942 \approx 5470.14\,\text{J mol}^{-1}$

Nearest integer:

$\boxed{5470}$

5. Comparison with stored correct answer

Stored answer = $2735$

My derived answer = $5470$

The stored answer appears to be exactly half of the correct value, which may arise if one incorrectly writes the reaction as

$HI \rightleftharpoons \frac12 H_2 + \frac12 I_2$

and then uses the equilibrium constant expression inconsistently with the given decomposition percentage. But the standard decomposition reaction is

$2HI \rightleftharpoons H_2 + I_2$

for which the correct value is

$\boxed{5470\,\text{J mol}^{-1}}$