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The reaction of cyanamide, [NH(2)CN(s)],...

The reaction of cyanamide, `[NH_(2)CN(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 enthalpy change for the reaction at `298 K`.
`NH_(2)CN(g)+3/2O_(2)(g) rarr N_(2)(g)+CO_(2)(g)+H_(2)O(l)`

Text Solution

Verified by Experts

`DeltaH = DeltaU + Deltan_(g) RT`
` DeltaU = -742.7 kJ mol^(-1), Deltan_(g) = 1 + 1 -3/2 = 1/2`
`therefore DeltaH = -742.7 kJ mol^(-1) + 1/2 xx 8.314 xx 10^(-3) kJ mol^(-1) K^(-1) xx 298 K`
`= - 742.7 + 1.2 `
` = -741.5 kJ mol^(-1)`
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