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If, S+O2 rarr SO2, Delta H =-298.2 kj ...

If, `S+O_2 rarr SO_2, Delta H =-298.2 kj`
`SO_2 + 1/2 O_2 rarr SO_3, Delta H=-98.7 kJ `
`SO_3 + H_2 O rarrr H_2SO_4 , Delta H =-130.2 kJ`
`H_2 + 1/2 O_2 rarr H_2 O, Delta H =-227.3 kJ`
Find `Delta H` for formation of `H_2SO_4` at 298 K.

A

`-754.4 kJ`

B

`+320.5 kJ`

C

`650.3 kJ`

D

`-433.7 kJ`

Text Solution

Verified by Experts

The correct Answer is:
A
Find `DeltaH` For `H_(2)+S+2O_(2)rarrH_(2)SO_(4)` Eq(i)+Eq(iv)+Eq(ii)+Eq(iii)
`DeltaH=(-298.2+248.3-98.7-130.2)` =-775.4KJ
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Given: (i) S+O_2 to SO_2, Delta H=-298.2 kJ (ii) SO_2 + 1/2 O_2 to SO_3, Delta H=-98.7 kJ (iii) SO_3 + H_2 O to H_2 SO_4, Delta H=-130.2 kJ (iv) H_2 + 1/2 O_2 to H_2 O, Delta H=-287.3 kJ , Then the enthalpy of formation of H_2 SO_4 at 298 K will be _________ .

If {:(SO_(2)+1/2O_(2) rarr SO_(3),DeltaH = -98.7 kJ),(SO_(3)+H_2O rarr H_(2)SO_(4),DeltaH = -130.2 kJ),(H_(2)+1/2O_(2) rarr H_2O , Delta H = -287.3 kJ),(S + H_2 + 2O_(2) rarr H_2SO_4, DeltaH = -814.4 kJ):} Then enthaply of SO_(2) at 298 K is:

S_R + O_2(g) rarr SO_2(g), Delta H =-71.1 kcal m S_M + O_2(g) rarr SO_2(g), Delta H =-71.7 kcal Thus Delta H for the consversion of S_R rarr S_M is

DeltaS_(sur) for H_(2)+1//2O_(2) rarr H_(2)O, Delta H = -280 KJ at 400 K is

If S+O_(2)toSO_(2),DeltaH=-298.2 " kJ" " mole"^(-1) SO_(2)+(1)/(2)O_(2)toSO_(3)DeltaH=-98.7 " kJ" " mole"^(-1) SO_(3)+H_(2)OtoH_(2)SO_(4),DeltaH=-130.2 " kJ" " mole"^(-1) H_(2)+(1)/(2)O_(2)toH_(2)SO_(4),DeltaH=-287.3 " kJ" " mole"^(-1) the enthlapy of formation of H_(2)SO_(4) at 298 K will be

C_(2)H_(2) + 5/2 O_2(2) rarr 2CO_(2) + H_(2)O , Delta H = -310 kcal C + O_(2) rarr CO_(2) , " "Delta H = -94 kcal H_(2) + 1/2 O_(2) rarr H_(2)O, " " Delta H = -68 kcal On the basis of the above equations, DeltaH_(f) (enthalpy of formation) of C_2H_2 will be :

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