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 \( A(g) + 3B(g) \rightarrow 2C(g) \) is in equilibrium, we can use the relationship between Gibbs free energy (\( \Delta G \)), enthalpy (\( \Delta H \)), and entropy (\( \Delta S \)). The equation we will use is:

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

At equilibrium, \( \Delta G = 0 \). Thus, we can set up the equation as follows:

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

Rearranging this equation gives us:

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

### Step 1: Convert the units of \( \Delta H \)

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

Since \( \Delta H \) is given in kilojoules, we need to convert it to joules:

\[ \Delta H = -100 \, \text{kJ} \times 1000 \, \text{J/kJ} = -100000 \, \text{J} \]

### Step 2: Substitute the values into the equation

Now we can substitute \( \Delta H \) and \( \Delta S \) into the equation for temperature:

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

### Step 3: Calculate the temperature

Now, perform the division:

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

Thus, the temperature at which the reaction will be in equilibrium is:

\[ \boxed{500 \, \text{K}} \]