Calculate the enthalpy change for the reaction between `CO_(2)` and `H_(2)O` to produce one mole of glucose `(C_(6)H_(12)O_(6))` . What would be enthalpy change for the production of `18` g of glucose ? The enthalpy of combustion of glucose is `2840 kJ mol^(-1)`.

Explanation

To calculate the enthalpy change for the reaction between \( CO_2 \) and \( H_2O \) to produce one mole of glucose \( C_6H_{12}O_6 \), we will use the enthalpy of combustion of glucose, which is given as \( -2840 \, \text{kJ/mol} \).

### Step 1: Write the combustion reaction of glucose
The combustion reaction of glucose can be represented as follows:

\[ C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O \]

The enthalpy change (\( \Delta H \)) for this reaction is \( -2840 \, \text{kJ/mol} \).

### Step 2: Reverse the combustion reaction
To find the enthalpy change for the formation of glucose from \( CO_2 \) and \( H_2O \), we need to reverse the combustion reaction:

\[ 6CO_2 + 6H_2O \rightarrow C_6H_{12}O_6 + 6O_2 \]

### Step 3: Change the sign of the enthalpy
When we reverse the reaction, the sign of the enthalpy change also reverses. Therefore, the enthalpy change for the formation of one mole of glucose is:

\[ \Delta H = +2840 \, \text{kJ/mol} \]

### Step 4: Calculate the enthalpy change for 18 g of glucose
Next, we need to calculate the enthalpy change for the production of 18 g of glucose. First, we need to find the number of moles of glucose in 18 g.

The molar mass of glucose \( C_6H_{12}O_6 \) is calculated as follows:

\[ \text{Molar mass of } C_6H_{12}O_6 = (6 \times 12.01) + (12 \times 1.008) + (6 \times 16.00) = 180.18 \, \text{g/mol} \]

Now, calculate the number of moles in 18 g:

\[ \text{Number of moles} = \frac{18 \, \text{g}}{180.18 \, \text{g/mol}} \approx 0.0999 \, \text{mol} \]

### Step 5: Calculate the enthalpy change for 0.0999 moles of glucose
Now we can calculate the enthalpy change for the production of 0.0999 moles of glucose:

\[ \Delta H_{\text{for 18 g}} = 0.0999 \, \text{mol} \times 2840 \, \text{kJ/mol} \approx 284.0 \, \text{kJ} \]

### Final Answer
The enthalpy change for the production of 18 g of glucose is approximately \( 284.0 \, \text{kJ} \).

---