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The enthalpy change DeltaH for the neutr...

The enthalpy change `DeltaH` for the neutralisation fo `1M HCI` by caustic potash in dilute solution at `298 K` is

A

`68 kJ`

B

`65 kJ`

C

`57.3kJ`

D

`50 kJ`

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To determine the enthalpy change (ΔH) for the neutralization of 1M HCl by caustic potash (KOH) in dilute solution at 298 K, we can follow these steps: ### Step 1: Understand the Reaction The neutralization reaction between hydrochloric acid (HCl) and potassium hydroxide (KOH) can be represented as: \[ \text{HCl (aq)} + \text{KOH (aq)} \rightarrow \text{KCl (aq)} + \text{H}_2\text{O (l)} \] ### Step 2: Identify the Type of Acid and Base Both HCl and KOH are strong acid and strong base, respectively. This means they will completely dissociate in solution. ...
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