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2 years ago

Glycerol boils at a higher temperature than water. What does this indicate about the attractive forces of glycerol?

2 answers:

7 0

The chemical formula for glycerol is HOCH2-CH(OH)-CH2OH while that of water is H-OH. The intermolecular forces in both water and glycerol are the strong H-bonds. However, since there are 3 OH groups/ glycerol molecule vs the one -OH group in water enables glycerol to form more H-bonds when compared to water.

The stronger attractive forces in glycerol causes an increase in its boiling point.

klio [65]2 years ago

6 0

Answer: Glycerol has more attractive forces as compared to water.

Explanation:

Boiling point is the temperature at which vapor pressure of the liquid becomes equal to atmospheric pressure.

Boiling point depends on the strength of inter molecular forces.

The molecules of glycerol $(HOCH_2CH(OH)CH_2OH)$ are more strongly bonded through hydrogen bonds as there are three OH groups.

But in water $(H_2O)$ , only on e OH group is present and thus extent of hydrogen bonding is less.

Thus we can conclude that glycerol has more attractive forces as compared to water.

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3 years ago

Calculate the enthalpy of the reaction

harkovskaia [24]

Answer : The enthalpy of the reaction is, -2552 kJ/mole

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given enthalpy of reaction is,

$4B(s)+3O_2(g)\rightarrow 2B_2O_3(s)$ $\Delta H=?$

The intermediate balanced chemical reactions are:

(1) $B_2O_3(s)+3H_2O(g)\rightarrow 3O_2(g)+B_2H_6(g)$ $\Delta H_A=+2035kJ$

(2) $2B(s)+3H_2(g)\rightarrow B_2H_6(g)$ $\Delta H_B=+36kJ$

(3) $H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)$ $\Delta H_C=-285kJ$

(4) $H_2O(l)\rightarrow H_2O(g)$ $\Delta H_D=+44kJ$

Now we have to revere the reactions 1 and multiple by 2, revere the reactions 3, 4 and multiple by 2 and multiply the reaction 2 by 2 and then adding all the equations, we get :

(when we are reversing the reaction then the sign of the enthalpy change will be change.)

The expression for enthalpy of the reaction will be,

$\Delta H=-2\times \Delta H_A+2\times \Delta H_B-6\times \Delta H_C-6\times \Delta H_D$

$\Delta H=-2(+2035kJ)+2(+36kJ)-6(-285kJ)-6(+44)$

$\Delta H=-2552kJ$

Therefore, the enthalpy of the reaction is, -2552 kJ/mole

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3 years ago