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Spontaneity

Determine Δ...

Question

Determine $Δ H$ for the following ration at $500 K$ and constant pressure:
$C O (g) + H_{2} O (g) \to C O_{2} (g) + H_{2} (g)$
Use the following data:
Substance $C_{p} (J / m o l K)$ $Δ_{f} H (298 K) (k J / m o l)$
$C O$ $29$ $- 110$
$H_{2} O$ $33$ $- 241$
$C O_{2}$ $37$ $- 393$
$H_{2}$ $29$ $0$

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Solution

$Δ H = Δ H_{p r o d u c t} - Δ H_{R e a c t a n t}$
$Δ H_{p r o d u c t} = 0 - 393 = - 393 k J m o l^{-}$
$Δ H_{r e a c t a n t} = - 110 - 241 = - 351 k J m o l^{-}$
$Δ H = - 393 - (- 351) = - 42 k J m o l^{-}$

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Δ H

for the following reaction at 500K and constant pressure :

C O (g) + H_{2} O (g) \to C O_{2} (g) + H_{2} (g)

use the following data :

Substance	$C_{P}$ ( J / mol K )	$Δ_{f} H$ ( 298K ) ( kJ / mol )
CO	29.12	$-$ 110.5
$H_{2} O$	33.58	$-$ 241.8
$C O_{2}$	37.11	$-$ 393.5
$H_{2}$	29.89	0.0

Q. A container contains

0.2 m o l

C O_{2}

0.1 m o l

H_{2}

and establish following equilibrium.

C O_{2} (g) + H_{2} (g) ⇌ C O (g) + H_{2} O (g)

. At equlibrium

H_{2} O

is found to be 25% by mol. The equilibrium constant

K p

for the reaction is

Δ H

k J

for the following reaction:

C (g) + O_{2} (g) \to C O_{2} (g)

H_{2} O (g) + C (g) \to C O (g) + H_{2} (g)

Δ H = + 131 k J

C O (g) + \frac{1}{2} O_{2} (g) \to {C O}_{2} (g)

Δ H = - 282 k J

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g)

Δ H = - 242 k J

Q. The value of

K_{c} = 4.24

800

K for the reaction

C O (g) + H_{2} O ⇌ C O_{2} (g) + H_{2} (g)

Calculate equilibrium concentrations of

C O_{2}, H_{2}, C O

H_{2} O

800

C O

H_{2} O

are present initially at concentration of

0.10

H_{2}

C O_{2}

at initial pressure of 10 atm and 20 atm respectively react at

1000

C O

H_{2} O

to attain equilibrium :

H_{2} (g) + C O_{2} (g) ⇌ H_{2} O (g) + C O (g)

. Calculate partial pressure of

H_{2} O

(nearest integer) (atm) at equilibrium if

K_{p}

1.60

1000

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