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The reaction of cyanamide (s), with diox...

The reaction of cyanamide (s), with dioxygen was carried out in a bomb calorimeter and `DeltaU` was found to be `-742.7` kJ `mol^(-1)` at 298 K. Calculate the enthalpy change for the reaction at 298 K.
`NH_(2)CN_((g))+(3)/(2)O_(2)(g)toN_(2(g))+CO_(2(g))+H_(2)O_(l)`

Text Solution

Verified by Experts

`NH_(2) CN_(g) +3/2 O_(2(g)) to N_(2(g)) +H_(2)O_(1) " "trianglen=2-5/2`
`triangleH=triangleU+trianglen RT`
`triangleH=-742+(-0.5) xx 8.314 xx 10^(-3) xx 298 " "triangleU="-742 kJ/mol"`
=-742.7 -1.238
`"=-743.94kJ/mol"^4`
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Knowledge Check

  • For the reaction CO_((g)) + (1)/(2) O_(2(g)), K_(1)//K_(c) is

    A
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    B
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  • If DeltaH_(f)^(0) " for" H_(2)O_(2(l)) and H_(2)O_((l)) " are " -188 kJ mol^(-1) and -286 kJ mol^(-1) , what will be the enthalpy change of the reaction 2H_(2)O_(2(l)) rarr 2H_(2)O_((l)) + O_(2(g)) = ?

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  • Which of the following holds good to the laws of thermodynamics for the reaction ? C_(2) H _(4 (g)) + 3O _(2 (g)) to 2 CO _(2(g)) + 2 H_(2) O _((l))

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