Isomers are compounds that have the same number and kinds of atoms but have a different arrangement of the atoms. The enthalpies of atom combination for several pairs of gaseos isomers are given here. For each pair, decide which has the strongest bonds.
a. CH3CH2CH2OH -4391.2kJ/molrxn
or CH3CHOH -4410.7 kJ/molrxn
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CH3
b. CH2=CHCH2CH3 -4604.9 kJ/molrxn
or CH3CH=CHCH3 -4611.9 kJ/molrxn
c. H2C=CHCH=CHCH3 -5243.3 kJ/molrxn
or H2C=CHCH2CH=CH2 -5213.5 kJ/molrxn
How would I go about this problem?
The CH3 in a is supposed to be under the second C in the compound.
To determine which pair of isomers has the strongest bonds, you would compare the enthalpies of atom combination for each pair. The lower the enthalpy of atom combination, the stronger the bond.
a. CH3CH2CH2OH or CH3CHOH
The enthalpy of atom combination for CH3CH2CH2OH is -4391.2 kJ/molrxn, and for CH3CHOH is -4410.7 kJ/molrxn. Since -4391.2 kJ/molrxn is a lower value than -4410.7 kJ/molrxn, CH3CH2CH2OH has stronger bonds.
b. CH2=CHCH2CH3 or CH3CH=CHCH3
The enthalpy of atom combination for CH2=CHCH2CH3 is -4604.9 kJ/molrxn, and for CH3CH=CHCH3 is -4611.9 kJ/molrxn. Since -4604.9 kJ/molrxn is a lower value than -4611.9 kJ/molrxn, CH2=CHCH2CH3 has stronger bonds.
c. H2C=CHCH=CHCH3 or H2C=CHCH2CH=CH2
The enthalpy of atom combination for H2C=CHCH=CHCH3 is -5243.3 kJ/molrxn, and for H2C=CHCH2CH=CH2 is -5213.5 kJ/molrxn. Since -5213.5 kJ/molrxn is a lower value than -5243.3 kJ/molrxn, H2C=CHCH2CH=CH2 has stronger bonds.
In summary, the pair with the stronger bonds for each case is:
a. CH3CH2CH2OH
b. CH2=CHCH2CH3
c. H2C=CHCH2CH=CH2
To determine which isomer has the strongest bonds, you need to compare the enthalpies of atom combination for each pair of isomers. The lower the enthalpy of atom combination, the stronger the bonds.
a. For the first pair, we have:
- CH3CH2CH2OH: -4391.2 kJ/molrxn
- CH3CHOH: -4410.7 kJ/molrxn
Since the enthalpy of atom combination for CH3CH2CH2OH is lower than that of CH3CHOH (-4391.2 kJ/molrxn < -4410.7 kJ/molrxn), CH3CH2CH2OH has stronger bonds.
b. For the second pair, we have:
- CH2=CHCH2CH3: -4604.9 kJ/molrxn
- CH3CH=CHCH3: -4611.9 kJ/molrxn
Here, the enthalpy of atom combination for CH2=CHCH2CH3 is lower than that of CH3CH=CHCH3 (-4604.9 kJ/molrxn < -4611.9 kJ/molrxn). Hence, CH2=CHCH2CH3 has stronger bonds.
c. For the third pair, we have:
- H2C=CHCH=CHCH3: -5243.3 kJ/molrxn
- H2C=CHCH2CH=CH2: -5213.5 kJ/molrxn
In this case, the enthalpy of atom combination for H2C=CHCH=CHCH3 is higher than that of H2C=CHCH2CH=CH2 (-5243.3 kJ/molrxn > -5213.5 kJ/molrxn). Therefore, H2C=CHCH=CHCH3 has weaker bonds compared to H2C=CHCH2CH=CH2.
By comparing the enthalpies of atom combination for each pair, you can determine which isomer in each case has the strongest bonds.