The reaction of cyanamide, NH2CN(s), with dioxygen was carried out in a bomb calorimeter, and ∆U was found to be –742.7 kJ mol–1 at 298 K. Calculate enthalpy change for the reaction at 298 K.

NH4CN(g) + \( \dfrac{3}{2}\)O2(g) → N2(g) + CO2(g) + H2O(l)

Asked by Abhisek | 1 year ago |  84

1 Answer

Solution :-

ΔH is given by,

\( \Delta H = \Delta U + \Delta n_{g}RT\)…………(1)

\( \Delta n_{g}\)​ = change in number of moles

\( \Delta U\) = change in internal energy

Here,

\( \Delta n_{g} = \sum n_{g}(product) – \sum n_{g}(reactant) \)

 ​​​​​​= (2 – 1.5) moles

 \( \Delta n_{g}\)​ = 0.5 moles

Here,

T =298K

\( \Delta U = -742.7 kJmol^{-1}\)

R  = \( 8.314\times 10^{-3}kJmol^{-1}K^{-1}\)

Now, from (1)

\( \Delta H = (-742.7 kJmol^{-1}) \)

\( + (0.5mol)(298K)\)

\( ( 8.314\times 10^{-3}kJmol^{-1}K^{-1})\)

= -742.7 + 1.2

\( \Delta H = -741.5 kJmol^{-1}\)

Answered by Pragya Singh | 1 year ago

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