Chemistry - Thermodynamics - I Question with Solution | TestHub

ChemistryThermodynamics - IHeat & Heat CapacityMedium2 minQB

ChemistryMediumnumerical

Passage / Comprehension

The evolution or absorption of heat, during solvation of a substance, is known as integral heat of solution ( $Δ H$ ). This $Δ H$ depends on pressure, temperature & moles of solvent for solvation of 1 mol substance. Given that : $Δ H = Δ H_{1} + Δ H_{2}$ Where $Δ H_{1}$ & $Δ H_{2}$ are differential heats of solution for solvent & solute respectively. Integral & differential heats of solution for $1 mol H_{2} SO_{4}$ in $H_{2} O$ at 1 bar and $2 5^{\circ} C$ is given below.

moles of $H_{2} O$	$Δ H$ (in cal)	$DH_{1}$ (in cal/mol)	$DH_{2}$ (in cal)
3	-11890	-1480	-7450
10	-16240	-x	-13910
$\infty$	-22990	0	-22990

moles of $H_{2} O$

$Δ H$ (in cal)

$DH_{1}$ (in cal/mol)

$DH_{2}$ (in cal)

-11890

-1480

-7450

-16240

-x

-13910

$\infty$

-22990

Calculate $∣ Δ_{r} H ∣$ (in kcal/mol) for the following reaction.

H₂SO₄ (3H₂) + 7H₂O(ℓ) → H₂SO₄ (10 H₂O)

Answer:

4.35

Solution:

$Δ_{r} H = - 16240 - (- 11890) = - 4350 cal / mol$

Stream:JEESubject:ChemistryTopic:Thermodynamics - ISubtopic:Heat & Heat Capacity

⏱ 2mℹ️ Source: QB

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