The reaction of cyanamide- NH 2 CN 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- NH 2 CN s -3-2 O 2 g - N 2 g - CO 2 g - H 2 O l

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Heat of Formation

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Question

The reaction of cyanamide, NH₂CN_(s),with dioxygen was carried out in a bomb calorimeter, and ΔU was found to be –742.7 kJ mol^–1at 298 K. Calculate enthalpy change for the reaction at 298 K.

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

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N H_{2} C N (s)

Δ U

- 742.7

k J m o l^{- 1}

298 K

298 K

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

(N H_{2} C N)

Δ U

- 742.7 {kJmol}^{- 1}

298 K

298 K

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

N H_{2} C N (s)

△ U

- 742.7 k J m o l^{- 1}

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

N H_{2} C N (s)

Δ U

- 742.7 k J {m o l}^{- 1}

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

Δ E

742.7 k J / m o l

N H_{2} C N (s)

298 K .

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

Δ H_{298}

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