Question

The heat of combustion of ethanol to give COz and water at constant pressure and 27°C is -327 kcal. How much heat is evolved in (cal) in combustion at the constant volume at 27°C?

Answer 1

GIVEN : -

• The heat of combustion of ethanol to give CO2 and water at constant pressure and 27°C is -327 kcal.

TO FIND:-

• Heat evolved in (cal) in combustion at the constant volume at 27°C.

SOLUTION:-

We know ,

$\implies \: ✪ \boxed{ \triangle \: H = \triangle \: E + nRT }$

So,

》 $\triangle$H = - 327 + 2 × 0.0821 × 300.

= -327 + 49.26

= -277.74

Hence,

$\implies \boxed{ \: \triangle \: H = - 277.74}</p><p></p><p>$

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HOPE IT HELPS :)

Answer 2

Answer- The above question is from the chapter 'Thermodynamics'.

Concept used: In chemical reaction,

reactants (g) → products (g)

$\tt PV_{r} = n_{r}RT\\\tt PV_{p} = n_{p}RT\\\tt P\triangle V = (n_{p} - n_{r})RT\\\tt P\triangle V= \triangle n_{g} RT$

$\boxed {\tt \implies \triangle H = \triangle U + \triangle n_g RT}$

Given question: The heat of combustion of ethanol to give CO₂ and water at constant pressure and 27°C is -327 kcal. How much heat is evolved in (cal) in combustion at the constant volume at 27°C ?

Answer: Temperature = 27°C = 27 + 273 = 300 K

ΔU = - 327 kcal = -327000 cal

Balanced chemical equation is as follows:-

CH₃CH₂OH (l) + 3O₂ (g) → 2CO₂ (g) + 3H₂O (l)

For gaseous components,

Number of moles in product side ($n_{p}$) = 2

Number of moles in reactant side ($n_{r}$) = 3

$\triangle n_g = n_p - n_r$ = 2 - 3 = -1

ΔH = ΔU + ΔnRT

ΔH = - 327 + [(-1) × 1.99 × 300]

ΔH = - 327000 - 597

ΔH = - 327597 cal

OR

ΔH = - 327.6 kcal

∴ Heat evolved in combustion of ethanol (in cal) = - 327597 cal.