`H^(+)(aq) + NaOH(aq) = Na^(+) + H_(2)O(l) ,DeltaH_(1)= -13690` cal
`HCN(aq) + NaOH(aq) to Na^(+) + CN^(-) + H_(2)O(l) ,DeltaH_(2) = -2900` cals
What is `DeltaH` value for HCN(aq) `=H^(+)(aq) + CN^(-)(aq)`

Options

A. 11400 cals
B. 10790 cals
C. 12500 cals
D. 9800 cals

Answer

To find the value of ΔH for the reaction \( \text{HCN(aq)} \rightarrow \text{H}^+(aq) + \text{CN}^-(aq) \), we can use Hess's law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps.

### Step-by-Step Solution:

1. **Write down the given reactions and their ΔH values:**
- Reaction 1:
\[ \text{H}^+(aq) + \text{NaOH}(aq) \rightarrow \text{Na}^+(aq) + \text{H}_2O(l), \quad \Delta H_1 = -13690 \text{ cal} \]
- Reaction 2:
\[ \text{HCN}(aq) + \text{NaOH}(aq) \rightarrow \text{Na}^+(aq) + \text{CN}^-(aq) + \text{H}_2O(l), \quad \Delta H_2 = -2900 \text{ cal} \]

2. **Invert Reaction 1 to get H⁺ on the product side:**
- Inverting Reaction 1 gives:
\[ \text{Na}^+(aq) + \text{H}_2O(l) \rightarrow \text{H}^+(aq) + \text{NaOH}(aq), \quad \Delta H = +13690 \text{ cal} \]

3. **Add the inverted Reaction 1 to Reaction 2:**
- Now, we add the inverted Reaction 1 to Reaction 2:
\[ (\text{Na}^+(aq) + \text{H}_2O(l) \rightarrow \text{H}^+(aq) + \text{NaOH}(aq)) + (\text{HCN}(aq) + \text{NaOH}(aq) \rightarrow \text{Na}^+(aq) + \text{CN}^-(aq) + \text{H}_2O(l)) \]

4. **Cancel out the common species:**
- On adding these two reactions, we can cancel out:
- \(\text{Na}^+(aq)\) from both sides.
- \(\text{NaOH}(aq)\) from both sides.
- \(\text{H}_2O(l)\) from both sides.
- This results in:
\[ \text{HCN}(aq) \rightarrow \text{H}^+(aq) + \text{CN}^-(aq) \]

5. **Calculate the total ΔH:**
- The total ΔH for the desired reaction is:
\[ \Delta H = \Delta H_1 + \Delta H_2 = 13690 \text{ cal} - 2900 \text{ cal} = 10790 \text{ cal} \]

### Final Answer:
\[ \Delta H = 10790 \text{ cal} \]