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Chemistry Level 3

Determine Δ U o \Delta U^{o} (in kJ) at 300 K 300K for the following reaction using the listed enthalpies of reaction.

Need to find out 4 C O ( g ) + 8 H 2 → 3 C H 4 ( g ) + C O 2 ( g ) + 2 H 2 O \large{4CO(g)+ 8H_{2} \rightarrow 3CH_{4}(g)+CO_{2}(g)+2H_{2}O}

Given

C g r a p h i t e + 1 2 O 2 ( g ) → C O ( g ) ; Δ H 1 o = − 110.5 k J C_{graphite}+\dfrac{1}{2} O_{2}(g) \rightarrow CO(g); \Delta H_{1}^{o}=-110.5 kJ

C O ( g ) + 1 2 O 2 ( g ) → C O 2 ( g ) ; Δ H 2 o = − 282.9 k J CO(g)+\dfrac{1}{2} O_{2}(g) \rightarrow CO_{2}(g); \Delta H_{2}^{o}=-282.9 kJ

H 2 ( g ) + 1 2 O 2 ( g ) → H 2 O ( l ) ; Δ H 3 o = − 285.8 k J H_{2}(g) + \dfrac{1}{2} O_{2}(g) \rightarrow H_{2}O(l); \Delta H_{3}^{o}=-285.8 kJ

C g r a p h i t e + 2 H 2 ( g ) → C H 4 ( g ) ; Δ H 4 o = − 74.8 k J C_{graphite}+2H_{2}(g) \rightarrow CH_{4}(g); \Delta H_{4}^{o}=-74.8 kJ

Find the Δ U o \Delta U^{o} (in kJ) at 300 K 300K for the reaction named Need to find out .

ORIGINAL

Try out my set Physical Chemistry For JEE Aspirants!


The answer is -727.44.

Md Zuhair by Md Zuhair , 20, India

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1 solution

Ayon Ghosh
Dec 23, 2017

Let us assume :

C g r a p h i t e + 1 2 O 2 ( g ) → C O ( g ) ; Δ H 1 o = − 110.5 k J . . . ( a ) C_{graphite}+\dfrac{1}{2} O_{2}(g) \rightarrow CO(g); \Delta H_{1}^{o}=-110.5 kJ ...(a)

C O ( g ) + 1 2 O 2 ( g ) → C O 2 ( g ) ; Δ H 2 o = − 282.9 k J . . . ( b ) CO(g)+\dfrac{1}{2} O_{2}(g) \rightarrow CO_{2}(g); \Delta H_{2}^{o}=-282.9 kJ...(b)

H 2 ( g ) + 1 2 O 2 ( g ) → H 2 O ( l ) ; Δ H 3 o = − 285.8 k J . . . ( c ) H_{2}(g) + \dfrac{1}{2} O_{2}(g) \rightarrow H_{2}O(l); \Delta H_{3}^{o}=-285.8 kJ...(c)

C g r a p h i t e + 2 H 2 ( g ) → C H 4 ( g ) ; Δ H 4 o = − 74.8 k J . . . ( d ) C_{graphite}+2H_{2}(g) \rightarrow CH_{4}(g); \Delta H_{4}^{o}=-74.8 kJ...(d)

Using Inspection Method ,The reaction which we desire possesses enthalpy equal to

− 3 ( a ) + ( b ) + 2 ( c ) + 3 ( d ) = − 747.4 K J m o l − 1 -3(a) + (b) + 2(c) + 3(d) = -747.4 KJ mol^{-1}

Since Δ U o \Delta U^{o} = Δ H o − Δ n g R T \Delta H^{o} - \Delta n_{g}RT where Δ n g = 4 − 12 = − 8 ; R = 8.314 ∗ 1 0 − 3 ; T = 300 \Delta n_{g} = 4-12 = -8 ; R = 8.314*10^{-3};T = 300

Δ U o ≈ − 727.4464 K J m o l − 1 \large \boxed{\Delta U^{o} \approx -727.4464 KJ mol^{-1} } .

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