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ANEK [815]
3 years ago
5

Conversion of 0.702 g of liquid water to steam at 100.0°C requires 1.56 kJ of heat. Calculate the molar enthalpy of evaporation

of water at 100.0°C.
Chemistry
1 answer:
kifflom [539]3 years ago
7 0

Answer:

The molar enthalpy of evaporation of water at 100.0°C is 40 kJ

Explanation:

Calculation of the moles of H_2O as:-

Mass = 0.702 g

Molar mass of H_2O = 18 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{0.702\ g}{18\ g/mol}

Moles= 0.039\ mol

Given that:- \Delta H=+1.56\ kJ

It means that when 0.039 moles of water undergoes evaporation, 1.56 kJ of energy is used

Also,

When 1 mole of water undergoes evaporation, \frac{1.56}{0.039}\ kJ of energy is used

<u>The molar enthalpy of evaporation of water at 100.0°C is 40 kJ</u>

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<u>Answer:</u> The final equation has hydroxide ions which indicate that the reaction has occurred in a basic medium.

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Redox reaction is defined as the reaction in which oxidation and reduction take place simultaneously.

The oxidation reaction is defined as the reaction in which a chemical species loses electrons in a chemical reaction. It occurs when the oxidation number of a species increases.

A reduction reaction is defined as the reaction in which a chemical species gains electrons in a chemical reaction. It occurs when the oxidation number of a species decreases.

The given redox reaction follows:

MnO_4^-(aq)+NO_2^-(aq)\rightarrow MnO_2(s)+NO_3^-(aq)

To balance the given redox reaction in basic medium, there are few steps to be followed:

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Oxidation half-reaction: NO_2^-+2OH^-\rightarrow NO_3^-+H_2O+2e^-

Reduction half-reaction: MnO_4^-+2H_2O+3e^-\rightarrow MnO_2+4OH^-

  • Multiply each half-reaction by the correct number in order to balance charges for the two half-reactions

Oxidation half-reaction: NO_2^-+2OH^-\rightarrow NO_3^-+H_2O+2e^-         ( × 3)

Reduction half-reaction: MnO_4^-+2H_2O+3e^-\rightarrow MnO_2+4OH^-             ( × 2)

The half-reactions now become:

Oxidation half-reaction: 3NO_2^-+6OH^-\rightarrow 3NO_3^-+3H_2O+6e^-

Reduction half-reaction: 2MnO_4^-+4H_2O+3e^-\rightarrow 2MnO_2+8OH^-

  • Add the equations and simplify to get a balanced equation

Overall redox reaction: 3NO_2^-+2MnO_4^-+H_2O\rightarrow 3NO_3^-+2MnO_2+2OH^-

As we can see that in the overall redox reaction, hydroxide ions are released in the solution. Thus, making it a basic solution

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