Question

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Enter your answer in the provided box. Acetylene burns in air according to the following equation: CH99)+0302 C029) + H206) AH = -1255.8 kJ I'xn Given sy of CO2(9)=–393.5 kJ/mol and sH of H2O(g)--241.8 kJ/mol, find all of C_H28). kJ/mol

Answer 1

Consider a given reaction , C2H2 (g) + 5 /2 O 2 (g)  $1585511199628_blob.png$2 CO 2(g) + H2O (g )

The standard enthalpy change of reaction is given as , $1585511199907_blob.png$ r H 0 = $1585511199640_blob.png$$1585511200053_blob.png$H 0f ( products ) -  $1585511199764_blob.png$$1585511200106_blob.png$H 0f ( reactants )

$1585511200212_blob.png$$1585511200338_blob.png$H 0f (products) =2 $1585511200466_blob.png$H 0f CO 2 (g) + $1585511200260_blob.png$H 0f H2O (g)

$1585511200696_blob.png$$1585511201880_blob.png$H 0f (products) = 2 (- 393.5 kJ mol -1 )  + ( - 241.8 kJ mol -1 )

$1585511200696_blob.png$$1585511201880_blob.png$H 0f (products) = - 1028.8 kJ mol -1

$1585511200732_blob.png$H 0f (reactants) = $1585511200746_blob.png$H 0f C2H2 (g) + 5/ 2 $1585511200725_blob.png$H 0f O 2 (g)

$1585511201886_blob.png$H 0f (reactants) = $1585511200760_blob.png$H 0f C2H2 (g) + 5/ 2 ( 0 ) kJ mol -1

i e $1585511201885_blob.png$H 0f (reactants) = $1585511200760_blob.png$H 0f C2H2 (g)

We have ,  $1585511201852_blob.png$ r H 0 = - 1255.8 kJ mol -1 & $1585511199907_blob.png$ r H 0 = $1585511199640_blob.png$$1585511200053_blob.png$H 0f ( products ) -  $1585511199764_blob.png$$1585511200106_blob.png$H 0f ( reactants )

$1585511201859_blob.png$ - 1255.8 kJ mol -1 =- 1028.8 kJ mol -1 - $1585511200760_blob.png$H 0f C2H2 (g)

$1585511201890_blob.png$ 1255.8 kJ mol -1 = 1028.8 kJ mol -1 + $1585511200760_blob.png$H 0f C2H2 (g)

$1585511201870_blob.png$$1585511200760_blob.png$H 0f C2H2 (g) = 1255.8 kJ mol -1 - 1028.8 kJ mol -1

$1585511200760_blob.png$H 0f C2H2 (g) = + 227.0 k J mol -1

ANSWER : $1585511200760_blob.png$H 0f C2H2 (g) = + 227.0 k J mol -1