For a hypothetical reaction `A(g) + 3B(g) to 2C(g). Delta H = -100 kJ and Delta S = -200 Jk^(-1)`. Then the temperature at which the reaction will be in equilibrium is

Options

A. 500 K
B. 480 K
C. 520 K
D. 510 K

Answer

To find the temperature at which the reaction will be in 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 up the equation as follows:

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

Rearranging this gives us:

\[ T = \frac{\Delta H}{\Delta S} \]

### Step 1: Substitute the values of \(\Delta H\) and \(\Delta S\)

Given:
- \(\Delta H = -100 \, \text{kJ}\)
- \(\Delta S = -200 \, \text{J/K}\)

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

\[ \Delta H = -100 \, \text{kJ} = -100 \times 10^3 \, \text{J} = -100000 \, \text{J} \]

### Step 2: Calculate the temperature \(T\)

Now substitute the values into the equation:

\[ T = \frac{-100000 \, \text{J}}{-200 \, \text{J/K}} \]

### Step 3: Simplify the equation

Calculating the right side:

\[ T = \frac{100000}{200} = 500 \, \text{K} \]

### Conclusion

The temperature at which the reaction will be in equilibrium is:

\[ T = 500 \, \text{K} \]