At 1 atm pressure, `DeltaS=75JK^(-1)mol^(-1),DeltaH=30kJ" "mol^(-1)`, the temperature of the reaction at equilibrium is

Options

A. 400 K
B. 330 K
C. 200 K
D. 110 K

Explanation

To find the temperature of the reaction 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 rearrange the equation to find the temperature \(T\):

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

This simplifies to:

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

Now, we can solve for \(T\):

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

### Step 1: Convert \(\Delta H\) to Joules
Given:
\[ \Delta H = 30 \text{ kJ/mol} = 30,000 \text{ J/mol} \]

### Step 2: Use the given \(\Delta S\)
Given:
\[ \Delta S = 75 \text{ J/K/mol} \]

### Step 3: Substitute values into the equation for \(T\)
Now we can substitute the values into the equation:

\[ T = \frac{30,000 \text{ J/mol}}{75 \text{ J/K/mol}} \]

### Step 4: Calculate \(T\)
Perform the calculation:

\[ T = \frac{30,000}{75} = 400 \text{ K} \]

### Final Answer
The temperature of the reaction at equilibrium is \(400 \text{ K}\).

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