For a reaction, `P+Q rarr R+S`. The value of `DeltaH^(@)` is `-"30 kJ mol"^(-1) and DeltaS" is "-"100 J K"^(-1)"mol"^(-1)`. At what temperature the reaction will be at equilibrium?

Options

A. `27^(@)C`
B. `52^(@)C`
C. `30^(@)C`
D. `45^(@)C`

Explanation

To find the temperature at which the reaction \( P + Q \rightleftharpoons R + S \) is at equilibrium, we can use the Gibbs free energy equation:

\[ \Delta G = \Delta H - T \Delta S \]

At equilibrium, \(\Delta G = 0\). Therefore, we can set the equation to:

\[ 0 = \Delta H - T \Delta S \]

Rearranging this gives us:

\[ \Delta H = T \Delta S \]

### Step 1: Substitute the given values
We know:
- \(\Delta H = -30 \, \text{kJ/mol}\)
- \(\Delta S = -100 \, \text{J/K/mol}\)

Before substituting, we need to convert \(\Delta H\) from kilojoules to joules:

\[ \Delta H = -30 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = -30000 \, \text{J/mol} \]

Now, substituting the values into the equation:

\[ -30000 \, \text{J/mol} = T \times (-100 \, \text{J/K/mol}) \]

### Step 2: Solve for T
We can simplify the equation:

\[ -30000 = -100T \]

Dividing both sides by -100:

\[ T = \frac{30000}{100} = 300 \, \text{K} \]

### Step 3: Convert Kelvin to Celsius
To convert the temperature from Kelvin to degrees Celsius, we use the formula:

\[ T(°C) = T(K) - 273 \]

Substituting the value we found:

\[ T(°C) = 300 \, \text{K} - 273 = 27 \, °C \]

### Final Answer
The temperature at which the reaction will be at equilibrium is:

\[ \boxed{27 \, °C} \]

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