The reaction of cyanamide, NH 2 CN (s) with oxygen was run in a bomb calorimeter and Δ U was found to be 742.24 kJ mol -1.The magnitude of Δ H 298 for the reaction NH 2 CN ( s )+(3/2) O 2 g arrow N 2( g )+ O 2( g )+ H 2 O (1) is - X kJ. Value of X is: (Rounded off to the nearest integer) [Assume ideal gases and R =8.314 J mol -1 K -1 ]

Question

The reaction of cyanamide, NH 2 CN (s) with oxygen was run in a bomb calorimeter and Δ U was found to be 742.24 kJ mol -1.The magnitude of Δ H 298 for the reaction NH 2 CN ( s )+(3/2) O 2 g arrow N 2( g )+ O 2( g )+ H 2 O (1) is - X kJ. Value of X is: (Rounded off to the nearest integer) [Assume ideal gases and R =8.314 J mol -1 K -1 ] (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

NH 2 CN ( s )+(3/2) O 2( g ) longrightarrow N 2( g )+ CO 2( g )+ H 2 O (ℓ) Δ n g =(1+1)-(3/2)=(1/2) Δ H =Δ U +Δ n g RT =-742.24+(1/2)× (8 . 314 × 298/1000) =-742.24+1.24 =-741kJ/mol X=741

NH2​CN(s)+23​O2​(g)⟶N2​(g)+CO2​(g)+H2​O(ℓ) Δng​=(1+1)−23​=21​ ΔH=ΔU+Δng​RT =−742.24+21​×10008.314×298​ =−742.24+1.24 =−741kJ/mol X=741

$N H_{2} CN (s) + \frac{3}{2} O_{2} (g) ⟶ N_{2} (g) + C O_{2} (g) + H_{2} O (ℓ)$
$Δ n_{g} = (1 + 1) - \frac{3}{2} = \frac{1}{2}$
$Δ H = Δ U + Δ n_{g} RT$
$= - 742.24 + \frac{1}{2} \times \frac{8.314 \times 298}{1000}$
$= - 742.24 + 1.24$
$= - 741 k J / m o l$
$X = 741$