Question

$\sf\red{The \: reaction \: of \: cyanamide, NH_2CN \: _(s_), \: with }$
$\sf\red{dioxygen \: was \: carried \: out \: in \: a \: bomb \: calorimeter, }$
$\sf\red{ and \:ΔU \: was \: found \: to \: be \: -742.7 \: kJ \: {mol}^{-1} \: at \: 298K.}$
$\sf\red{ Calulate \: enthalpy \: change \: for \: the \: reaction \: 298K.}$

$\bf{NH_2CN \: (g) + \frac{3}{2} O_2 \: (g) → N_2 (g)+ CO_2 (g)+H_2O (l)}$


​

Answer 1

Answer:

For given reaction, Δn=1+1−1.5=0.5.

For given reaction, Δn=1+1−1.5=0.5.ΔH=ΔU+ΔngRT

For given reaction, Δn=1+1−1.5=0.5.ΔH=ΔU+ΔngRT=−742.7+0.5×8.314×10−3×298=−742.7+1.2=−741.5 kJ mole−1

Answer 2

Answer:

For given reaction, Δn=1+1−1.5=0.5.

ΔH=ΔU+Δn

g

RT

=−742.7+0.5×8.314×10

−3

×298=−742.7+1.2=−741.5 kJ mole

−1