Calculate enthalpy change for the change 8 S g - S 8 g - given that H - 2 S -2 g - rightarrow 2 H - g -2 S - g - Welta H -239-0 - k - cal - mol -1H2S 2 g rightarrow 2 H - g - S - g - Delta H -175-0 - k - cal -mol-1 A- -508-0 k calB- -512-0 k calC- 508-0 k calD- -512-0 k cal

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Standard XII

Chemistry

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

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- 512.0 k cal

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508.0k cal

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-508.0 k cal

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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 $8 \times (- 64) = - 512 k c a l$

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Similar questions

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

The enthalpies for the following reactions at 25°C are given.
$\frac{1}{2} H_{2} (g) + \frac{1}{2} O_{2} (g) \to O H (g); △ H = 10.06 K c a l$
$H_{2} (g) \to 2 H (g); △ H = 104.18 K c a l$
$O_{2} (g) \to 2 O (g); △ H = 118.32 K c a l$

Calculate the OH bond energy in the O - H group

Q. Find the heat change in the reaction

N H_{3} (g) + H C l (g) \to N H_{4} C l (s)

from the following data

N H_{3} (g) + a q \to N H_{3} (a q),

Δ H = - 8.4 K . C a l .

H C l (g) + a q \to H C l (a q),

Δ H = - 17.3 K . C a l .

N H_{3} (a q) + H C l (a q) \to N H_{4} C l (a q), Δ H = - 12.5 K . C a l s .

N H_{4} C l (s) + a q \to N H_{4} C l (a q), Δ H = + 3.9 K . C a l .

Q. The enthalpy change for the reaction

H_{2} (g) + C_{2} H_{4} (g) \to C_{2} H_{6} (g)

is:

The bond energies are,

H - H = 103, C - H = 99, C - C = 80

and

C = C = 145

K c a l m o l^{- 1}

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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 $8 \times (- 64) = - 512 k c a l$