The reaction of cyanamide NH 2 CN ( s ) with oxygen was carried out in a bomb calorimeter and Δ q v at 300 K was measured to be -750 kJ / mol. The value of Δ H per mole of NH 2 CN ( s ) in the given reaction is

Question

The reaction of cyanamide NH 2 CN ( s ) with oxygen was carried out in a bomb calorimeter and Δ q v at 300 K was measured to be -750 kJ / mol. The value of Δ H per mole of NH 2 CN ( s ) in the given reaction is (A) -741.75 kJ / mol (B) -748.75 kJ / mol (C) -752.75 kJ / mol (D) -750 kJ / mol

Tardigrade Admin · Accepted Answer

The combustion reaction is: NH 2 CN ( s )+(3/2) O 2( g ) arrow N 2( g )+ CO 2( g )+ H 2 O ( l ) Δ n=(1/2) Δ H=Δ E+Δ n R T=-750+(1/2) × 8.314 × 10-3 × 300 or Δ H=-750+0.15 × 8.314=-748.75 kJ / mol

Q. The reaction of cyanamide $N H_{2} CN (s)$ with oxygen was carried out in a bomb calorimeter and $Δ q_{v}$ at $300 K$ was measured to be $- 750 k J / m o l$ . The value of $Δ H$ per mole of $N H_{2} CN (s)$ in the given reaction is

-741.75 kJ / mol

-748.75 kJ / mol

-752.75 kJ / mol

-750 kJ / mol

The combustion reaction is:

$N H_{2} CN (s) + \frac{3}{2} O_{2} (g) \to N_{2} (g) + C O_{2} (g) + H_{2} O (l)$

$Δ n = \frac{1}{2}$

$Δ H = Δ E + Δ n RT = - 750 + \frac{1}{2} \times 8.314 \times 1 0^{- 3} \times 300$

or $Δ H = - 750 + 0.15 \times 8.314 = - 748.75 k J / m o l$

Q. The reaction of cyanamide NH2​CN(s) with oxygen was carried out in a bomb calorimeter and Δqv​ at 300K was measured to be −750kJ/mol. The value of ΔH per mole of NH2​CN(s) in the given reaction is