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DeltaS(sur) for H(2)+1//2O(2) rarr H(2)O...

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

A

700J/g/K

B

700KJ/mol/K

C

700 J/mol/K

D

0.7J/mol/K

Text Solution

Verified by Experts

The correct Answer is:
C
`Delta S_(sys) = -Delta S_("surr")`
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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:

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Given that - 2C(s)+2O_(2)(g)rarr 2CO_(2)(g) Delta H = -787 KJ H_(2)(g)+ 1//2O_(2)(g)rarr H_(2)O(l) Delta =-286 KJ C_(2)H_(2)(g)+(5)/(2)O_(2)(g)rarr H_(2)O(l) Delta H=-1310 KJ Heat of formation of acetylene is :-

Calculate in kJ for the following reaction : C(g) + O_(2)(g) rarr CO_(2)(g) Given that, H_(2)O(g) + C(g) + H_(2)(g) , Delta H = +131 kJ CO(g) + 1/2 O_(2)(g) rarr CO_(2)(g), " " Delta H = -242 kJ H_(2)(g) + 1/2 O_(2)(g) rarr H_(2)O(g), " "DeltaH = -242 kJ

With the help of thermochemical equations given below, determine Delta_(r )H^(Θ) at 298 K for the following reaction: C("graphite")+2H_(2)(g) rarr CH_(4)(g),Delta_(r )H^(Θ) = ? C("graphite")+O_(2)(g) rarr CH_(2)(g), Delta_(r )H^(Θ) = -393.5 kJ mol^(-1) ...(1) H_(2)(g) +1//2O_(2)(g) rarr H_(2)O(l) , Delta_(r )H^(Θ) = -285.8 kJ mol^(-1) ...(2) CO_2(2)(g)+2H_(2)O(l) rarr CH_(4)(g)+2O_(2)(g) , Delta_(r )H^(Θ) = +890.3 kJ mol^(-1) ...(3)

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