The reaction of cyanamide, NH₂CN_(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

$N H_{2} C N_{(S)} + \frac{3}{2} O_{2 (g)} \to N_{2 (g)} + O_{2 (g)} + H_{2} O_{(I)} i s$ ------------- kJ. (Rounded off to the nearest integer)

[Assume ideal gases and R = 8.314 J mol^-1 K^-1]

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Vishal Baghel | Contributor-Level 10

3 weeks ago

$Δ U = - 742.24 k J / m o l e Δ H_{298} = ?$

NH₂CN (s) + $\frac{3}{2}$ O₂ (g) ® N₂ (g) + O₂ (g) + H₂O (l)

= $- 742.24 + \frac{1}{2} \times \frac{8.314}{1000} \times 298$

$- 742.24 + 1.238 = - 741.002 \approx - 741 k J / m o l$

So, magnitude of $Δ H$ = 741 kJ/mol.

<math> <mrow> <mi>Δ</mi> <mi>U</mi> <mo>=</mo> <mo>−</mo> <mn>7</mn> <mn>4</mn> <mn>2</mn> <mo>.</mo> <mn>2</mn> <mn>4</mn> <mtext> </mtext> <mi>k</mi> <mi>J</mi> <mo>/</mo> <mi>m</mi> <mi>o</mi> <mi>l</mi> <mi>e</mi> <mtext> </mtext> <mtext> </mtext> <mi>Δ</mi> <msub> <mrow> <mi>H</mi> </mrow> <mrow> <mn>2</mn> <mn>9</mn> <mn>8</mn> </mrow> </msub> <mo>=</mo> <mo>? </mo> </mrow> </math> <!- [endif]-><!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1025" DrawAspect="Content" ObjectID="_1819618851"> </o:OLEObject></xml><! [endif]->NH2CN (s) + <math> <mrow> <mfrac> <mrow> <mn>3</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> </math> <!- [if gte vml 1]><v:shape id="_x0000_i1026" type="#_x0000_t75" style='width:11pt;height:29pt' o:ole=""> <v:imagedata src="file://C://Users/VISHAL~1.BAG/AppData/Local/Temp/msohtmlclip1/01/clip_image003.wmz" o:title=""/></v:shape><! [endif]-><!- [if !vml]-><img src="file://C://Users/VISHAL~1.BAG/AppData/Local/Temp/msohtmlclip1/01/clip_image004.png" width="15" height="39"><!- [endif]-><!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1026" DrawAspect="Content" ObjectID="_1819618852"> </o:OLEObject></xml><! [endif]->O2 (g) ® N2 (g) + O2 (g) + H2O (l)= <math> <mrow> <mo>−</mo> <mn>7</mn> <mn>4</mn> <mn>2</mn> <mo>.</mo> <mn>2</mn> <mn>4</mn> <mo>+</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>×</mo> <mfrac> <mrow> <mn>8</mn> <mo>.</mo> <mn>3</mn> <mn>1</mn> <mn>4</mn> </mrow> <mrow> <mn>1</mn> <mn>0</mn> <mn>0</mn> <mn>0</mn> </mrow> </mfrac> <mo>×</mo> <mn>2</mn> <mn>9</mn> <mn>8</mn> </mrow> </math> <!- [endif]-><!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1027" DrawAspect="Content" ObjectID="_1819618853"> </o:OLEObject></xml><! [endif]-> <math> <mrow> <mo>−</mo> <mn>7</mn> <mn>4</mn> <mn>2</mn> <mo>.</mo> <mn>2</mn> <mn>4</mn> <mo>+</mo> <mn>1</mn> <mo>.</mo> <mn>2</mn> <mn>3</mn> <mn>8</mn> <mo>=</mo> <mo>−</mo> <mn>7</mn> <mn>4</mn> <mn>1</mn> <mo>.</mo> <mn>0</mn> <mn>0</mn> <mn>2</mn> <mo>≈</mo> <mo>−</mo> <mn>7</mn> <mn>4</mn> <mn>1</mn> <mi>k</mi> <mi>J</mi> <mo>/</mo> <mi>m</mi> <mi>o</mi> <mi>l</mi> </mrow> </math> So, magnitude of <math> <mrow> <mi>Δ</mi> <mi>H</mi> </mrow> </math> = 741 kJ/mol.

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