Given that- H2g-I2s- 2HIg- H-51-9 KJ-mole H2g-I2g- 2HIg- H-9-2 KJ-moleThe heat of reaction of I2s- I2g is-

The correct option is B +61.1 KJ/mole(i) H_2(g)+I_2(s)→ 2HI(g); Δ H=51.9 KJ/mole #160; #160; .....(1)(ii) H_2(g)+I_2(g)→ 2HI(g); Δ H= 9.2 KJ/m ...

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Byju's Answer

Standard XII

Chemistry

Enthalpy

Given that, ...

Question

Given that,

$H_{2} (g) + I_{2} (s) \to 2 H I (g); Δ H = 51.9 K J / m o l e$
$H_{2} (g) + I_{2} (g) \to 2 H I (g); Δ H = - 9.2 K J / m o l e$

The heat of reaction of $I_{2} (s) \to I_{2} (g)$ is:

-61.1 KJ/mole

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+61.1 KJ/mole

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+31.1 KJ/mole

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50.1 KJ/mole

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Solution

The correct option is B +61.1 KJ/mole
(i) $H_{2} (g) + I_{2} (s) \to 2 H I (g); Δ H = 51.9$ KJ/mole .....(1)

(ii) $H_{2} (g) + I_{2} (g) \to 2 H I (g); Δ H = - 9.2$ KJ/mole ......(2)

(iii) $2 H I (g) \to H_{2} (g) + I_{2} (g); Δ H = 9.2$ KJ/mole ......(3)

Adding (i) & (iii), we get:

$I_{2} (s) \to I_{2} (g) Δ H = 51.9 + 9.2$ KJ/mole $= 61.1$ KJ/mole

Hence, option B is correct.

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Similar questions

$H_{2} (g) + I_{2} (g) \to 2 H I (g); Δ H = 51.9$ kJ

According to the above reaction, the heat of formation of $H I$ (in kJ/mol) is:

Q. For a given reaction,

H_{2} (g) + I_{2} (g) ⇌ 2 H I (g)

4.5

moles each of

H_{2}

and

I_{2}

are heated in a sealed

10 L

vessel. At equilibrium,

3

mole of

H I

were found. What is the value of equilibrium constant?

Q. For the reaction

H_{2} (g) + I_{2} (g) ⇌ 2 H I (g), K_{c} = 47.6

, If the initial number of moles of each reactant and product is 1 mole, then in equilibrium:

Q. 1 mole of

H_{2}

, 2 moles of

I_{2}

and 3 moles of HI were taken in 1 litre flask. The equilibrium constant of the reaction:

H_{2 (g)} + I_{2 (g)} ⇌ 2 H I_{(g)}

is 50 at

330^{o} C

. The concentration of HI at equilibrium is:

Q. 1 mole of

H_{2}, 2

moles of

I_{2}

and 3 moles of

H I

were taken in

1

litre flask. The equilibrium constant of the reaction

H_{2 (g)} + I_{2 (g)} ⇌ 2 H I_{(g)}

50

330^{o} C .

The concentration of

H I

at equilibrium is: (moles.lit

^{- 1}

)

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Given that, H2(g)+I2(s)→2HI(g);ΔH=51.9KJ/moleH2(g)+I2(g)→2HI(g);ΔH=−9.2KJ/moleThe heat of reaction of I2(s)→I2(g) is:

The correct option is B +61.1 KJ/mole(i) H2(g)+I2(s)→2HI(g);ΔH=51.9 KJ/mole .....(1)(ii) H2(g)+I2(g)→2HI(g);ΔH=−9.2 KJ/mole ......(2)(iii) 2HI(g)→H2(g)+I2(g);ΔH=9.2 KJ/mole ......(3)Adding (i) & (iii), we get:I2(s)→I2(g) ΔH=51.9+9.2 KJ/mole =61.1 KJ/moleHence, option B is correct.

Given that,

$H_{2} (g) + I_{2} (s) \to 2 H I (g); Δ H = 51.9 K J / m o l e$
$H_{2} (g) + I_{2} (g) \to 2 H I (g); Δ H = - 9.2 K J / m o l e$

The heat of reaction of $I_{2} (s) \to I_{2} (g)$ is: