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-BARSIC- [3]
3 years ago
11

nonpolar solutes dissolve more readily in nonpolar solvents than polar solvents. Describe the interactions occuring between the

solute and the solvent
Chemistry
1 answer:
earnstyle [38]3 years ago
8 0

Answer:

The interactions are known as London dispersion forces.

Explanation:

It is widely known that the electron density distribution in polar and nonpolar molecules is not uniform and not static. In most cases, the electron density in any molecule will adjust to create a spontaneous dipole. In the presence of nonpolar solute and nonpolar solvent, the spontaneous dipoles have a propagating effect on the interaction. If molecule A produces a spontaneous dipole, then the electron density in the other molecule B will react by creating an opposite spontaneous dipole. The dipoles have interactive forces known as the London dispersion forces.The magnitudes of these forces increase with increase in the surface area of the molecule. Therefore, molecules with larger surface areas will have stronger London dispersion forces than those with smaller surface areas.

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Determine the enthalpy change for the reaction 2C(s) + 2H2O(g) → CH4(g) + CO2(g) using the following:
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Answer : The enthalpy change for the reaction is, 97.7 kJ

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 main chemical reaction is,

2C(s)+3H_2O(g)\rightarrow CH_4(g)+CO_2(g)    \Delta H=?

The intermediate balanced chemical reaction will be,

(1) C(s)+H_2O(g)\rightarrow CO(g)+H_2(g)     \Delta H_1=131.3kJ

(2) CO(g)+H_2O(g)\rightarrow CO_2(g)+H_2(g)    \Delta H_2=41.2kJ

(3) CH_4(g)+H_2O(g)\rightarrow 2H_2(g)+CO(g)    \Delta H_3=206.1kJ

Now we are multiplying reaction 1 by 2 and reversing reaction 3 and then adding all the equations, we get :

(1) 2C(s)+2H_2O(g)\rightarrow 2CO(g)+2H_2(g)     \Delta H_1=2\times 131.3kJ=262.6kJ

(2) CO(g)+H_2O(g)\rightarrow CO_2(g)+H_2(g)    \Delta H_2=41.2kJ

(3) 2H_2(g)+CO(g)\rightarrow CH_4(g)+H_2O(g)    \Delta H_3=-206.1kJ

The expression for enthalpy of main reaction will be,

\Delta H=\Delta H_1+\Delta H_2+\Delta H_3

\Delta H=(262.6)+(41.2)+(-206.1)

\Delta H=97.7kJ

Therefore, the enthalpy change for the reaction is, 97.7 kJ

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Explanation:

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Answer:

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Explanation:

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