The heat of neutralisation of a strong base and a strong acid is 13.7 kcal. The heat released when 0.7 mole HBr solution is added to 0.35 mole of KOH is

Options

A. 5.425 kcal
B. 13.7 kcal
C. `4.795 kcal
D. 8.795 kcal

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Reaction
The neutralization reaction between a strong acid (HBr) and a strong base (KOH) can be represented as:
\[ \text{HBr} + \text{KOH} \rightarrow \text{KBr} + \text{H}_2\text{O} \]

### Step 2: Identify Moles of Reactants
We are given:
- Moles of HBr = 0.7 moles
- Moles of KOH = 0.35 moles

### Step 3: Determine the Limiting Reagent
In a neutralization reaction, the limiting reagent is the one that will be completely consumed first. Here, KOH is the limiting reagent because we have fewer moles of KOH (0.35 moles) compared to HBr (0.7 moles).

### Step 4: Calculate Moles of Water Produced
Since 1 mole of KOH reacts with 1 mole of HBr to produce 1 mole of water, the amount of water produced will be equal to the moles of KOH:
- Moles of water produced = 0.35 moles

### Step 5: Calculate Heat Released
The heat of neutralization for the reaction between a strong acid and a strong base is given as 13.7 kcal per mole of water produced. Therefore, for 0.35 moles of water produced, the heat released can be calculated as:
\[ \text{Heat released} = \text{Moles of water} \times \text{Heat of neutralization} \]
\[ \text{Heat released} = 0.35 \, \text{moles} \times 13.7 \, \text{kcal/mole} \]
\[ \text{Heat released} = 4.795 \, \text{kcal} \]

### Final Answer
The heat released when 0.7 moles of HBr solution is added to 0.35 moles of KOH is **4.795 kcal**.

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