MyQuestionIcon

12

You visited us 12 times! Enjoying our articles? Unlock Full Access!

Gibbs Free Energy & Spontaneity

Calculate the...

Question

Calculate the standard free energy change for the combustion of glucose at $298 K$ , using the given data.
$C_{6} H_{12} O_{6} + 6 O_{2} (g) ⟶ 6 {C O}_{2} (g) + 6 H_{2} O$
$Δ H^{o} = - 2820 k J {m o l}^{- 1}; Δ S^{o} = 210 J K^{- 1} {m o l}^{- 1}$

A

- 1882.58 k J {m o l}^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B

- 2882.58 k J {m o l}^{- 1}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

C

- 3882.58 k J {m o l}^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

D

- 4882.58 k J {m o l}^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $- 2882.58 k J {m o l}^{- 1}$
For standard free energy change we have
$Δ G = Δ H - T * Δ S$
$Δ G = - 2820000 - 298 * 210 = - 2882580 J m o l^{- 1}$
$Δ G = - 2882.58 K J m o l^{- 1}$

Suggest Corrections

thumbs-up

0

Similar questions

{Δ H}_{f}^{o}

C_{6} H_{12} O_{6} (s)

from the following data:

Δ H_{c o m b .}

C_{6} H_{12} O_{6} (s) = - 2816 k J {m o l}^{- 1}

{Δ H}_{f}^{o}

{C O}_{2} (g) = - 393.5 J {m o l}^{- 1}

{Δ H}_{f (H_{2} O)}^{o} = - 285.9 k J {m o l}^{- 1}

.

Q. For the reaction

2 N O C l (g) ⇌ 2 N O (g) + {C l}_{2} (g)

calculate the standard equilibrium constant at

298 K

. Given that the values of

Δ H^{o}

Δ S^{o}

of the reaction at

298 K

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

122 J K {m o l}^{- 1}

Q. The disachharide

α

- maltose can be hydrolysed to glucose according to the equation (only magnitude in nearest integer in kj/mol)

C_{12} H_{22} O_{11} (a q) + H_{2} O (l) ⟶ 2 C_{6} H_{12} O_{6} (a q)

Using the following values, calculate the standard enthalpy change in this reaction:

Δ_{f} H^{o} (H_{2} O, l) = - 285.85 k J

{m o l}^{- 1}

Δ_{f} H^{o} (C_{6} H_{12} O_{6}, a q) = - 1263.1 k J

{m o l}^{- 1}

Δ_{f} H^{o} (C_{12} H_{22} O_{11}, a q) = - 2238.3.1 k J

{m o l}^{- 1}

Q. The equilibrium constant for the reaction

{C O}_{2} (g) + H_{2} (g) ⇌ C O (g) + H_{2} O (g)

298 K

7

. Calculate the value of standard free energy change.

(R = 8.314 J K^{- 1} {m o l}^{- 1})

Q. The dissociation constant of

{N H}_{4} O H

25^{o} C

K_{b} = X \times 10^{- 6}

Δ H^{o}

Δ S^{o}

for the given changes are as follows:

{N H}_{4} + H^{+} ⇌ {N H}_{4}^{+}; Δ H^{o} = - 52.2 k J {m o l}^{- 1}; Δ S^{o} = + 1.67 J K^{- 1} {m o l}^{- 1}

H_{2} O ⇌ H^{+} + O H^{-}; Δ H^{o} = 56.6 k J {m o l}^{- 1}; Δ S^{o} = - 78.2 J K^{- 1} {m o l}^{- 1}

. Value of X is=5Y+2. What is the value of Y?

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Gibbs Free Energy

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Gibbs Free Energy & Spontaneity

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon