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 l Given the enthalpies of formation- H f 0 C 57 H 104 O 6- s -70870 kJ - mol- H f 0 H 2 O - l -285-8 kJ - mol- H f 0 CO 2- g -393-5 kJ - molA- -37-98B- -30-2C- -33-4D- -40-4

The correct option is A 37.98 Δ H_f^0 =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.

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

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

- 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

Q. 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

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

N H_{3} (g)

reacts with

C u O (s)

according to the following reaction:

3 C u O (s) + 2 N H_{3} (g) \to 3 C u (s) + N_{2} (g) + 3 H_{2} O (l)

Standard enthalpies of formation of

N H_{3} (g), C u O (s)

and

H_{2} O (l)

are

- 46 {kJ mol}^{- 1}, - 155 {kJ mol}^{- 1}, and - 285 {kJ mol}^{- 1}

respectively. The magnitude of heat liberated (in

kJ

) when

2.72 g

N H_{3} (g)

is reacted completely with

C u O (s)

according to the reaction given above when all the reactants and products are in their standard state is:

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.

Q. a) The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are

- 890.3 k J^{- 1}

- 393.5 k J^{- 1}

and

- 285.8 k J^{- 1}

respectively. Calculate the enthalpy of formation of

{C H}_{4} (g)

b) Write the relationship between Gibb's energy, equilibrium constant and change in enthalpy.

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The fat C57H104O6 (s) is metabolized via the following reaction. calculate the energy (KJ) liberated when 1g of this fat reacts. C57H104O6(s)+80O2(g)→57CO2(g)+52H2O(l) Given the enthalpies of formation: ΔH0f(C57H104O6,s)=−70870 kJ/mol ΔH0f(H2O,l)=−285.8 kJ/mol ΔH0f(CO2,g)=−393.5 kJ/mol

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

The correct option is A
- 37.98

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