`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 the given reactions and their enthalpy changes (ΔH values).

### 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. **Identify the target reaction:**
- The target reaction we want to find ΔH for is:
\[ \text{HCN}(aq) \rightarrow \text{H}^+(aq) + \text{CN}^-(aq) \]

3. **Manipulate the given reactions:**
- We can rearrange Reaction 1 to isolate H⁺:
\[ \text{Na}^+(aq) + \text{H}_2O(l) \rightarrow \text{H}^+(aq) + \text{NaOH}(aq) \quad \Delta H = +13690 \text{ cal} \]
- This means we reverse the reaction, which changes the sign of ΔH.

4. **Combine the reactions:**
- Now, we can combine Reaction 2 and the reversed Reaction 1:
\[ \text{HCN}(aq) + \text{NaOH}(aq) \rightarrow \text{Na}^+(aq) + \text{CN}^-(aq) + \text{H}_2O(l) \quad \Delta H_2 = -2900 \text{ cal} \]
\[ \text{Na}^+(aq) + \text{H}_2O(l) \rightarrow \text{H}^+(aq) + \text{NaOH}(aq) \quad \Delta H = +13690 \text{ cal} \]

5. **Add the ΔH values:**
- The overall ΔH for the target reaction can be calculated as follows:
\[ \Delta H = \Delta H_2 + \Delta H_1 \text{ (reversed)} \]
\[ \Delta H = -2900 \text{ cal} + 13690 \text{ cal} \]
\[ \Delta H = 10790 \text{ cal} \]

### Final Answer:
The ΔH value for the reaction \(\text{HCN(aq)} \rightarrow \text{H}^+(aq) + \text{CN}^-(aq)\) is **10790 cal**.