The molar entropies of HI g - H g and I g at 298 K are 206-5-114-6- and 180-7 J mole -1 K -1 respectively- Using the - G 0 given below- Calculate the bond energy of HI- g - H g - I g - - G 0-271-8 kJH A- 282-4 kj - moleB- 298-3 kJ - moleC- 100 kJ - moleD- 32-17 kJ - mole

The correct option is B 298.3 kJ/mole HI_(g) → HI_(g) + I_(g) Δ G^0 = Δ G TΔ S^0 Δ S^0 = ∑ S^0 products ∑ S^0 Reactants = (114.6 + 180.7) (206.5) = 2 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Hess' Law

The molar ent...

Question

The molar entropies of $H I (g), H (g)$ and $I (g)$ at $298 K$ are $206.5, 114.6$ , and $180.7 J m o l e^{- 1} K^{- 1}$ respectively. Using the $Δ G^{0}$ given below,Calculate the bond energy of HI.

$H (g) \to H (g) + I (g)$ ; $Δ G^{0} = 271.8 k J$

282.4 kj/mole

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

298.3 kJ/mole

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

32.17kJ/mole

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

100 kJ/mole

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

Open in App

Solution

The correct option is B
298.3 kJ/mole

$H I_{(g)} \to H I_{(g)} + I_{(g)}$

$Δ G^{0} = Δ G - T Δ S^{0}$

$Δ S^{0} = \sum S^{0} p r o d u c t s - \sum S^{0} R e a c t a n t s$

= $(114.6 + 180.7) - (206.5) = 295.3 - 206.5 = 88.8$

$Δ G^{0} = Δ H - T Δ S$

$Δ H = Δ G^{0} + T Δ S = 27.18 + 298 \times 88.8 = 271.8 + 26462.8 J o u l e s$

= $271.8 + 26.4628 = 298.3 k J / m o l e$

Suggest Corrections

Similar questions

The molar entropies of $H I (g), H (g)$ and $I (g)$ at $298 K$ are $206.5, 114.6$ , and $180.7 J m o l e^{- 1} K^{- 1}$ respectively. Using the $Δ G^{0}$ given below,Calculate the bond energy of HI.

$H (g) \to H (g) + I (g)$ ; $Δ G^{0} = 271.8 k J$

Q. The entropies of

H_{2} (g)

and

H (g)

are

130.6

and

114.6

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

respectively at

298 K

. Using the data given below calculate the bond energy of

H_{2}

(in

k J / m o l

H_{2} (g) \to 2 H (g); Δ G^{0} = 406.6 k J

Q. The entropies of

H_{2} (g)

and

H (g)

are

130.6

and

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

respectively at

298 K

. Using the data given below calculate the bond energy of

H_{2}

(in kJ/mol).

Q. The standard enthalpy of combustion at

25^{\circ} C

of hydrogen, cyclohexene

(C_{6} H_{10})

and cyclohexane

(C_{6} H_{12})

are

- 241, - 3800

and

- 3920 K J / m o l e

respectively. Calculate the heat of hydrogenation of cyclohexene.

Enthalpy of hydrogenation is the enthalpy change when one mole of unsaturated compound is converted to saturated compound. $Δ_{H} h y d r o g e n a t i o n$ helps to calculate resonance energy.

The standard enthalpy of combustion at $25^{\circ} C$ of hydrogen, cyclohexene and cyclohexane are -300, -3600 and -3770 KJ/mole respectively. Calculate the heat of hydrogenation of cyclohexene.

Join BYJU'S Learning Program

The molar entropies of HI(g),H(g) and I(g) at 298K are 206.5,114.6, and 180.7 J mole−1 K−1 respectively. Using the Δ G0 given below,Calculate the bond energy of HI. H(g) → H(g) + I(g); Δ G0 = 271.8 kJ

The molar entropies of $H I (g), H (g)$ and $I (g)$ at $298 K$ are $206.5, 114.6$ , and $180.7 J m o l e^{- 1} K^{- 1}$ respectively. Using the $Δ G^{0}$ given below,Calculate the bond energy of HI.

$H (g) \to H (g) + I (g)$ ; $Δ G^{0} = 271.8 k J$