For a reversible reaction `A hArr B`. Find `(log_(10)K)/(10)` at `2727^(@)`C temperature
Given
`Delta_(r)H^(0)` = - 54.07 kJ `mol^(-1)`
`Delta_(r)S^(0)` = 10 `JK^(-1)`
R = 8.314 `JK^(-1) mol^(-1)`

Answer

To solve the problem, we will use the relationship between the Gibbs free energy change (ΔG°), the enthalpy change (ΔH°), the entropy change (ΔS°), and the equilibrium constant (K) for the reaction. The steps are as follows:

### Step 1: Convert the temperature from Celsius to Kelvin
Given the temperature is 2727°C, we convert it to Kelvin using the formula:
\[ T(K) = T(°C) + 273 \]
\[ T = 2727 + 273 = 3000 \, K \]

**Hint:** Remember to always convert Celsius to Kelvin by adding 273.

### Step 2: Write the relationship between ΔG°, ΔH°, and ΔS°
The Gibbs free energy change is related to the enthalpy and entropy changes by the equation:
\[ \Delta G° = \Delta H° - T \Delta S° \]

**Hint:** This equation shows how Gibbs free energy is affected by enthalpy and entropy.

### Step 3: Relate ΔG° to the equilibrium constant K
The relationship between Gibbs free energy and the equilibrium constant is given by:
\[ \Delta G° = -RT \ln K \]
Where R is the gas constant.

**Hint:** This equation connects thermodynamic properties with the equilibrium constant.

### Step 4: Equate the two expressions for ΔG°
By equating the two expressions for ΔG°, we have:
\[ \Delta H° - T \Delta S° = -RT \ln K \]

**Hint:** This step combines the two equations to find a relationship involving K.

### Step 5: Solve for ln K
Rearranging the equation gives:
\[ \ln K = \frac{\Delta H° - T \Delta S°}{-RT} \]

**Hint:** Isolate ln K to express it in terms of ΔH°, ΔS°, T, and R.

### Step 6: Convert ln K to log₁₀ K
Using the conversion between natural logarithm and base 10 logarithm:
\[ \ln K = 2.303 \log_{10} K \]
Thus,
\[ \log_{10} K = \frac{\ln K}{2.303} \]

**Hint:** Remember the conversion factor between natural log and log base 10.

### Step 7: Substitute values into the equation
Now we substitute the values into the equation:
- ΔH° = -54.07 kJ/mol = -54070 J/mol (convert kJ to J)
- ΔS° = 10 J/K
- R = 8.314 J/K·mol
- T = 3000 K

So we have:
\[ \log_{10} K = \frac{\Delta H° - T \Delta S°}{-RT \cdot 2.303} \]

### Step 8: Calculate log₁₀ K
Substituting the values:
\[ \log_{10} K = \frac{-54070 - (3000)(10)}{- (8.314)(3000) \cdot 2.303} \]
\[ \log_{10} K = \frac{-54070 - 30000}{- (8.314)(3000) \cdot 2.303} \]
\[ = \frac{-84070}{- (8.314)(3000) \cdot 2.303} \]

Calculating the denominator:
\[ = (8.314)(3000) \cdot 2.303 \approx 57440.1 \]

Thus,
\[ \log_{10} K \approx \frac{84070}{57440.1} \approx 1.46 \]

### Step 9: Find log₁₀ K / 10
Finally, we need to find:
\[ \frac{\log_{10} K}{10} \approx \frac{1.46}{10} = 0.146 \]

### Final Answer
Thus, the value of \(\frac{\log_{10} K}{10}\) at 2727°C is approximately \(0.146\).

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