$H_{2} O (l) ⟶ H_{2} O (s); Δ H = - 6.01 k J$
$Δ H$ is the heat of freezing, (solidification) of water.If true enter 1 else 0
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Solution

The correct option is A 1
$H_{2} O (l) ⟶ H_{2} O (s); Δ H = - 6.01 k J$
The enthalpy change $Δ H$ is the heat of freezing, (solidification) of water. It is the enthalpy change associated with conversion of one mole of liquid into solid state at its freezing point.
It is always negative as heat is liberated when the intermolecular forces between constituent particles of solid are formed.

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