The standard change in Gibbs energy for the reaction: `H_2O hArr H^+ +OH^-` at `25^@C` is :

Options

A. `100 kJ`
B. `-80 kJ`
C. `80 k J`
D. `-100 kJ`

Answer

To find the standard change in Gibbs energy (ΔG°) for the reaction:

\[ H_2O \rightleftharpoons H^+ + OH^- \]

at \( 25^\circ C \), we can use the following thermodynamic relationship:

\[ \Delta G = \Delta G^\circ + RT \ln K \]

At equilibrium, ΔG = 0, hence:

\[ 0 = \Delta G^\circ + RT \ln K \]

Rearranging this gives:

\[ \Delta G^\circ = -RT \ln K \]

### Step 1: Identify the values of R, T, and K

- **R** (universal gas constant) = \( 8.314 \, \text{J K}^{-1} \text{mol}^{-1} \)
- **T** (temperature in Kelvin) = \( 25^\circ C = 298 \, \text{K} \)
- **K** (equilibrium constant for the dissociation of water) = \( K_w = [H^+][OH^-] = 10^{-14} \)

### Step 2: Calculate \( \ln K \)

Since \( K = 10^{-14} \):

\[ \ln K = \ln(10^{-14}) = -14 \ln(10) \]

Using \( \ln(10) \approx 2.303 \):

\[ \ln K \approx -14 \times 2.303 \approx -32.242 \]

### Step 3: Substitute the values into the equation for ΔG°

Now substitute R, T, and \( \ln K \) into the equation:

\[ \Delta G^\circ = - (8.314 \, \text{J K}^{-1} \text{mol}^{-1}) \times (298 \, \text{K}) \times (-32.242) \]

### Step 4: Perform the calculation

Calculating the product:

\[ \Delta G^\circ = 8.314 \times 298 \times 32.242 \]

Calculating \( 8.314 \times 298 \):

\[ 8.314 \times 298 \approx 2477.572 \, \text{J mol}^{-1} \]

Now multiply by \( 32.242 \):

\[ \Delta G^\circ \approx 2477.572 \times 32.242 \approx 79892.3 \, \text{J mol}^{-1} \]

### Step 5: Convert to kilojoules

To convert from joules to kilojoules:

\[ \Delta G^\circ \approx 79.8923 \, \text{kJ mol}^{-1} \approx 80 \, \text{kJ mol}^{-1} \]

### Final Answer

The standard change in Gibbs energy for the reaction at \( 25^\circ C \) is approximately:

\[ \Delta G^\circ \approx 80 \, \text{kJ mol}^{-1} \]

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