The fat C 57 H 104 O 6 s is metabolized via the following reaction- calculate the energy KJ liberated when 1 g of this fat reacts- C 57 H 104 O 6 s -80 O 2 g - 57 CO 2 g -52 H 2 O I Given the enthalpies of formation-VDelta H - 0- f C -57- H -104 O - 6- - S -70870 kJ - mol 1Delta H - 0- f H - 2 - O - l -285-8 kJ - mol lDelta H - 0- f CO - 2- g -393-5 kJ - mol A- -37-98B- -40-4C- -33-4D- -30-2

The correct option is A 37.98Δ H^0_f = 57 × ( 393.5) + 52 × ( 285.8) ( 70870) = 33578.9 kJ/mol = 33578.9/884 = 37.98 KJ ...

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Byju's Answer

Standard XII

Chemistry

Law of Mass Action

The fat C 57 ...

Question

The fat $C_{57} H_{104} O_{6} (s)$ is metabolized via the following reaction. calculate the energy (KJ) liberated when 1g of this fat reacts.
$C_{57} H_{104} O_{6} (s) + 80 O_{2} (g) \to 57 C O_{2} (g) + 52 H_{2} O (I)$
Given the enthalpies of formation:
$Δ H_{f}^{0} (C_{57} H_{104} O_{6}, s) = - 70870 k J / m o l Δ H_{f}^{0} (H_{2} O, l) = - 285.8 k J / m o l Δ H_{f}^{0} (C O_{2}, g) = - 393.5 k J / m o l$

- 37.98

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Solution

The correct option is A - 37.98
$Δ H_{f}^{0} = 57 \times (- 393.5) + 52 \times (- 285.8) - (- 70870) = - 33578.9 k J / m o l = - \frac{33578.9}{884} = - 37.98 K J$

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Similar questions

The fat $C_{57} H_{104} O_{6}$ (s) is metabolized via the following reaction.
calculate the energy (KJ) liberated when 1g of this fat reacts.

$C_{57} H_{104} O_{6} (s) + 80 O_{2} (g) \to 57 C O_{2} (g) + 52 H_{2} O (l)$

Given the enthalpies of formation:

$Δ H_{f}^{0} (C_{57} H_{104} O_{6}, s) = - 70870$ kJ/mol
$Δ H_{f}^{0} (H_{2} O, l) = - 285.8$ kJ/mol
$Δ H_{f}^{0} (C O_{2}, g) = - 393.5$ kJ/mol

Q. The fat, glyceryl trioleate, is metabolized via the following reaction. Given the enthaplies of formation, calculate the energy (kJ) liberated when 1.00 g of this fat reacts.
(Atomic weights :

C = 12.01, H = 1.008, O = 16.00

C_{57} H_{104} O_{6} (s) + 80 O_{2} (g) ⟶ 57 C O_{2} (g) + 52 H_{2} O (I)

Δ H C_{57} H_{107} O_{6} = - 70870 k J / m o l e

Δ H H_{2} O (I) = - 285.8 k J / m o l e

Δ H C O_{2} (g) = - 393.5 k J / m o l e

Q. The reaction of 1 mole of benzene take place at

298 K

and

1 a t m

as-

C_{6} H_{6} (l) + \frac{15}{2} O_{2} (g) \to 6 C O_{2} (g) + 3 H_{2} O (l)

in this reaction

C O_{2} (g)

and

H_{2} O (l)

are produced and 3267.0

k J

heat is liberated. calculate the

Δ_{f} H^{o}

for benzene.

Δ_{f} H^{o}

for

C O_{2} (g)

and

H_{2} O (l)

are

- 393.5 k J m o l^{- 1}

and

- 285.83 k J m o l^{- 1}

respectively.

Determine enthalpy of formation for $H_{2} O_{2} (l),$ using the listed enthalpies of reactions:

$N_{2} H_{4} (l) + 2 H_{2} O_{2} (l) \to N_{2} (g) + 4 H_{2} O (l);$

$△_{r} H_{1}^{o} = - 818 k J / m o l$

$N_{2} H_{4} (l) + O_{2} (g) \to N_{2} (g) + 2 H_{2} O (l);$ $△_{r} H_{2}^{o} = - 622 k J / m o l$

$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); △_{f} H_{3}^{o} = - 285 k J / m o l$

Q. Calculate the enthalpy of the reaction:

S n O_{2} (s) + 2 H_{2} (g) ⟶ S n (s) + 2 H_{2} O (l)

given that, enthalpies of formation of

S n O_{2} (s)

and

H_{2} O (l)

are

- 580.7 k J

and

- 285.8 k J

respectively.

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The correct option is A - 37.98ΔH0f=57×(−393.5)+52×(−285.8)−(−70870)=−33578.9 kJ/mol=−33578.9884=−37.98 KJ

The correct option is A - 37.98
$Δ H_{f}^{0} = 57 \times (- 393.5) + 52 \times (- 285.8) - (- 70870) = - 33578.9 k J / m o l = - \frac{33578.9}{884} = - 37.98 K J$