
If \[\Delta H_r^o\] for \[{H_2}{O_2}\] and \[{H_2}O\] are \[ - 188KJ/mol\] and \[ - 286KJ/mol\], what will be the enthalpy change of the reaction? \[2{H_2}{O_2} \to 2{H_2}O + {O_2}\]
A.\[ - 196KJ/mol\]
B.\[ - 494KJ/mol\]
C.\[ - 146KJ/mol\]
D.\[ - 98KJ/mol\]
Answer
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Hint: As we know that the enthalpy of fusion is also called the heat of fusion. The enthalpy of a fusion of a substance indicates the change in the enthalpy or heat developed from providing the energy. The particular amount of energy of a substance to change the phase of a material from solid state to liquid phase at a constant pressure. The enthalpy of the reaction can be found by adding the internal energy with the product of pressure and the volume and the enthalpy is represented as ‘H’.
Complete answer:
According to the question, \[\Delta H_r^o\] for \[{H_2}{O_2}\] and \[{H_2}O\] are \[ - 188KJ/mol\] and \[ - 286KJ/mol\] respectively.
Here, two moles of hydrogen peroxide are decomposed into two moles of water and one mole of oxygen.
The enthalpy change of the reaction can be find out by using the equation,
\[\Delta {{\rm H}_{rxn}} = \sum {\Delta {{\rm H}_{product}} - } \sum {\Delta {{\rm H}_{product}}} \]
Substitute the given values in above equation, will get;
\[\Delta {{\rm H}_{rxn}} = \left( {2 \times - 286} \right) - \left( {2 \times - 188} \right)\]
\[ = - 572 + 376\]
On addition we get,
\[ = - 196kJ/mol\]
Hence, option (A) is correct.
The change in enthalpy of the reaction is not equal to\[ - 494KJ/mol\]. Hence, option (B) is incorrect.
The enthalpy change is not equal to \[ - 146KJ/mol\]. Hence, option (C) is incorrect.
The change in enthalpy of the given reaction is not equal to \[ - 98KJ/mol\]. Hence, option (D) is incorrect.
Hence, option (A) is correct.
Note:
The change in enthalpy of reaction is calculated by subtracting the enthalpy of the product. Here, two moles of hydrogen peroxide are decomposed into two moles of water and one mole of oxygen. The quantity of heat energy for the conversion of the definite mass of a solid at its melting point to the liquid is known as enthalpy of fusion. Normally, the chemical reaction will undergo the enthalpy change. Sometimes, we can directly calculate the enthalpy change by subtracting the values of enthalpy of products and the reactant.
Complete answer:
According to the question, \[\Delta H_r^o\] for \[{H_2}{O_2}\] and \[{H_2}O\] are \[ - 188KJ/mol\] and \[ - 286KJ/mol\] respectively.
Here, two moles of hydrogen peroxide are decomposed into two moles of water and one mole of oxygen.
The enthalpy change of the reaction can be find out by using the equation,
\[\Delta {{\rm H}_{rxn}} = \sum {\Delta {{\rm H}_{product}} - } \sum {\Delta {{\rm H}_{product}}} \]
Substitute the given values in above equation, will get;
\[\Delta {{\rm H}_{rxn}} = \left( {2 \times - 286} \right) - \left( {2 \times - 188} \right)\]
\[ = - 572 + 376\]
On addition we get,
\[ = - 196kJ/mol\]
Hence, option (A) is correct.
The change in enthalpy of the reaction is not equal to\[ - 494KJ/mol\]. Hence, option (B) is incorrect.
The enthalpy change is not equal to \[ - 146KJ/mol\]. Hence, option (C) is incorrect.
The change in enthalpy of the given reaction is not equal to \[ - 98KJ/mol\]. Hence, option (D) is incorrect.
Hence, option (A) is correct.
Note:
The change in enthalpy of reaction is calculated by subtracting the enthalpy of the product. Here, two moles of hydrogen peroxide are decomposed into two moles of water and one mole of oxygen. The quantity of heat energy for the conversion of the definite mass of a solid at its melting point to the liquid is known as enthalpy of fusion. Normally, the chemical reaction will undergo the enthalpy change. Sometimes, we can directly calculate the enthalpy change by subtracting the values of enthalpy of products and the reactant.
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