Calculate enthalpy change for the change 8Sg- S8g- given thatH2S2g- 2Hg - 2Sg - - H - 239-0 k cal mol-1H2S2g- 2Hg - Sg- - H - 175-0 k cal mol-1

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

Enthalpy

Calculate ent...

Question

Calculate enthalpy change for the change $8 S_{(g)} \to S_{8 (g)^{'}}$ given that
$H_{2} S_{2 (g)} \to 2 H_{(g)} + 2 S_{(g)}, Δ H = 239.0 k c a l m o l^{- 1} H_{2} S_{2 (g)} \to 2 H_{(g)} + S_{(g)}, Δ H = 175.0 k c a l m o l^{- 1}$

+ 512.0 k cal

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

- 512.0 k cal

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

508.0k cal

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

-508.0 k cal

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

Open in App

Solution

The correct option is B
- 512.0 k cal

$Δ H_{S - S} + 2 Δ H_{H - S} = 239$ $2 Δ H_{H - S} = 175$
Hence, $Δ H_{S - S} = 239 - 175 = 64 k c a l m o l^{- 1}$
Then, $Δ H$ for $8 S_{(g)} \to S_{g (g)}$ is $8 \times (- 64) = - 512 k c a l$

Suggest Corrections

Similar questions

Calculate enthalpy change for the change $8 S_{(g)} \to S_{8 (g)^{'}}$ given that
$H_{2} S_{2 (g)} \to 2 H_{(g)} + 2 S_{(g)}, Δ H = 239.0 k c a l m o l^{- 1} H_{2} S_{2 (g)} \to 2 H_{(g)} + S_{(g)}, Δ H = 175.0 k c a l m o l^{- 1}$

Q. For a chemical reaction the values of

Δ H

and

Δ S

at 400 K are -10 K cal mol

^{- 1}

and 20 cal. deg

^{- 1}

mol

^{- 1}

respectively. Calculate the value of

Δ G

of the reaction.

$K_{c}$ for the reaction, $F e^{2 +} + Z n \to Z n^{2 +} + F e$ is $10^{23}$ , The standard free energy change is

Q. The enthalpy change for a given reaction at

298 K

- x c a l / m o l

. If the reaction occurs spontaneously at

298 K

, the entropy change at that temperature:

Join BYJU'S Learning Program

Calculate enthalpy change for the change 8S(g)→S8(g)′ given that H2S2(g)→2H(g)+2S(g),ΔH=239.0 k cal mol−1H2S2(g)→2H(g)+S(g),ΔH=175.0 k cal mol−1

The correct option is B - 512.0 k cal ΔHS−S+2ΔHH−S=239 2ΔHH−S=175 Hence, ΔHS−S=239−175=64 kcal mol−1 Then, ΔH for 8S(g)→Sg(g) is 8×(−64)=−512k cal

Calculate enthalpy change for the change $8 S_{(g)} \to S_{8 (g)^{'}}$ given that
$H_{2} S_{2 (g)} \to 2 H_{(g)} + 2 S_{(g)}, Δ H = 239.0 k c a l m o l^{- 1} H_{2} S_{2 (g)} \to 2 H_{(g)} + S_{(g)}, Δ H = 175.0 k c a l m o l^{- 1}$

The correct option is B
- 512.0 k cal

$Δ H_{S - S} + 2 Δ H_{H - S} = 239$ $2 Δ H_{H - S} = 175$
Hence, $Δ H_{S - S} = 239 - 175 = 64 k c a l m o l^{- 1}$
Then, $Δ H$ for $8 S_{(g)} \to S_{g (g)}$ is $8 \times (- 64) = - 512 k c a l$