Calculate - H at 358 K for the reaction-Fe2O3s- 3H2g - 2Fes-3H2OlGiven that - - H298- -33-29 kJ mol-1 and CP for Fe2O3s- Fes- H2Ol and H2g are 103-8- 25-1- 75-3- and 28-8 J-K mol-

Reaction taking place is, Fe_2O_3(s)+ 3H_2(g) → 2Fe(s)+3H_2O(l) for this, C_p = (C_p)_product (C_p)_reactant putting the values, Δ C_P=[2× 25.1 + 3× 75.3] [1 ...

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Standard XI

Chemistry

Heat of Reaction

Calculate Δ H...

Question

Calculate $Δ H$ at $358 K$ for the reaction,
$F e_{2} O_{3} (s) + 3 H_{2} (g) \to 2 F e (s) + 3 H_{2} O (l)$

Given that , $Δ H_{298} = - 33.29 k J m o l^{- 1}$ and $C_{P} f o r F e_{2} O_{3} (s), F e (s), H_{2} O (l)$ and $H_{2} (g)$ are $103.8, 25.1, 75.3, a n d 28.8 J / K m o l$ .

- 28.136 k J / m o l

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- 58.136 k J / m o l

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28 .136 k J / m o l

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58.136 k J / m o l

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Solution

The correct option is A $- 28.136 k J / m o l$
Reaction taking place is,
$F e_{2} O_{3} (s) + 3 H_{2} (g) \to 2 F e (s) + 3 H_{2} O (l)$
for this,
$△ C_{p} = (C_{p})_{p r o d u c t} - (C_{p})_{r e a c t a n t}$
putting the values,
$Δ C_{P} = [2 \times 25.1 + 3 \times 75.3] - [103.8 + 3 \times 28.8]$
$= 85.9 J / K m o l$

We have, from Kirchoff equation,
$\frac{Δ H_{2} - Δ H_{1}}{T_{2} - T_{1}} = Δ C_{P}$
putting the values,
$\frac{Δ H_{358} - (- 33290)}{358 - 298} = 85.9$
$Δ H_{358} = - 28136 J / m o l$
$= - 28.136 k J / m o l$

Theory:
Kirchoff’s Equation:
Kirchhoff's Law describes the enthalpy of a reaction's variation with temperature changes. The overall enthalpy of the reaction will change if the increase in the enthalpy of products and reactants is different.

Kirchoff’s equation can be written as:

If $C_{p, m}$ is given as a function of temperature,

$Δ_{r} H_{2} = Δ_{r} H_{1}$ + $\int_{T_{1}}^{T_{2}} Δ_{r} C_{p, m} d T$

Where $T_{1}$ and $T_{2}$ are in Kelvin $(K)$

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Similar questions

Q. Calculate

Δ H

358 K

for the reaction,

F e_{2} O_{3} (s) + 3 H_{2} (g) \to 2 F e (s) + 3 H_{2} O (l)

Given that ,

Δ H_{298} = - 33.29 k J m o l^{- 1}

and

C_{P} f o r F e_{2} O_{3} (s), F e (s), H_{2} O (l)

and

H_{2} (g)

are

103.8, 25.1, 75.3, a n d 28.8 J / K m o l

Q. The

Δ H

value at 360 K for the reaction:

F e_{2} O_{3 (s)} + 3 H_{2 (g)} \to 2 F e_{(s)} + 3 H_{2} O_{(l)}

, given that

Δ H_{300} = - 33.29 k J m o l^{- 1}

and

C_{p}

for

F e_{2} O_{3 (s)}, F e_{(s)}, H_{2} O (l)

and

H_{2 (g)}

are

103.8, 25.1, 75.3

and

28.8

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

, is:

Q. Calculate

Δ H

358 K

for the reaction,

F e_{2} O_{3} (s) + 3 H_{2} (g) \to 2 F e (s) + 3 H_{2} O (l)

Given that ,

Δ H_{298} = - 33.29 k J m o l^{- 1}

and

C_{P} f o r F e_{2} O_{3} (s), F e (s), H_{2} O (l)

and

H_{2} (g)

are

103.8, 25.1, 75.3, a n d 28.8 J / K m o l

Q. Calculate, the

Δ H

85^{o} C

for the reaction:

{F e}_{2} O_{3} (s) + 3 H_{2} ⟶ 2 F e (s) + 3 H_{2} O (l)

The data are:

Δ H_{298}^{o} = - 33.29 k J / m o l

and

Substance	${F e}_{2} O_{3} (s)$	$F e (s)$	$H_{2} O (l)$	$H_{2} (g)$
$C_{P}^{o} (J / K m o l)$	$103.8$	$25.1$	$75.3$	$28.8$

Q. Calculate, the

Δ G^{o}

85^{o} C

for the reaction:

F e_{2} O_{3} (s) + 3 H_{2} (g) \to 2 F e (s) + 3 H_{2} O

The data are:

Δ H_{298}^{o} = - 35 k J / m o l

Δ S_{358}^{o} = 0.095 k J m o l^{- 1} K^{- 1}

S u b s t a n c e F e_{2} O_{3} (s) F e (s) H_{2} O (l) H_{2} (g)

C_{p}^{o} (J K^{- 1} m o l - 1) 103257529

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Heat of Reaction

Standard XI Chemistry

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Calculate ΔH at 358 K for the reaction, Fe2O3(s)+3H2(g) → 2Fe(s)+3H2O(l) Given that , ΔH298=−33.29 kJ mol−1 and CP for Fe2O3(s), Fe(s), H2O(l) and H2(g) are 103.8, 25.1, 75.3, and 28.8 J/K mol.

Calculate $Δ H$ at $358 K$ for the reaction,
$F e_{2} O_{3} (s) + 3 H_{2} (g) \to 2 F e (s) + 3 H_{2} O (l)$

Given that , $Δ H_{298} = - 33.29 k J m o l^{- 1}$ and $C_{P} f o r F e_{2} O_{3} (s), F e (s), H_{2} O (l)$ and $H_{2} (g)$ are $103.8, 25.1, 75.3, a n d 28.8 J / K m o l$ .