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acrylonitrile, an important starting material for the manufacture of pl…

Question

acrylonitrile, an important starting material for the manufacture of plastics and synthetic rubber, is made from propene: 2c3h6 + 2nh3 + 3o2 → 2h2c=ch−c≡n + 6h2o. use average bond energies to estimate the energy change for the acrylonitrile synthesis. attempts: 0 of 6 used save for later etextbook and media submit answer current attempt in progress view policies

Explanation:

Step1: Identify bonds broken and formed

First, list all the bonds broken and formed in the reaction $2C_3H_6 + 2NH_3+3O_2
ightarrow2H_2C = CH - C\equiv N+6H_2O$. For reactants, in $C_3H_6$ there are $C - C$ and $C - H$ bonds, in $NH_3$ there are $N - H$ bonds and in $O_2$ there is $O = O$ bond. For products, in $H_2C=CH - C\equiv N$ there are $C = C$, $C\equiv N$ and $C - H$ bonds, and in $H_2O$ there are $O - H$ bonds.

Step2: Look up bond - energy values

Use standard bond - energy values. For example, $C - C$ bond energy is around $347$ kJ/mol, $C - H$ is around $413$ kJ/mol, $N - H$ is around $391$ kJ/mol, $O = O$ is around $498$ kJ/mol, $C = C$ is around $614$ kJ/mol, $C\equiv N$ is around $891$ kJ/mol, and $O - H$ is around $463$ kJ/mol.

Step3: Calculate energy of bonds broken

In $2C_3H_6$: There are $2$ moles of $C - C$ bonds and $12$ moles of $C - H$ bonds. In $2NH_3$: There are $6$ moles of $N - H$ bonds. In $3O_2$: There are $3$ moles of $O = O$ bonds.
The energy of bonds broken $E_{broken}=2\times347+12\times413 + 6\times391+3\times498$
$E_{broken}=694+4956+2346 + 1494$
$E_{broken}=9490$ kJ.

Step4: Calculate energy of bonds formed

In $2H_2C = CH - C\equiv N$: There are $2$ moles of $C = C$ bonds, $2$ moles of $C\equiv N$ bonds and $6$ moles of $C - H$ bonds. In $6H_2O$: There are $12$ moles of $O - H$ bonds.
The energy of bonds formed $E_{formed}=2\times614+2\times891+6\times413+12\times463$
$E_{formed}=1228+1782+2478+5556$
$E_{formed}=11044$ kJ.

Step5: Calculate energy change

The energy change $\Delta E=E_{broken}-E_{formed}$
$\Delta E = 9490 - 11044=- 1554$ kJ.

Answer:

$-1554$ kJ