Answer
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Hint: In Chemistry, Chemical equilibrium is a condition in which the concentration of reactants and products does not change with respect to time and the rate of forward reaction becomes equal to the rate of backward reaction. The rate is determined with the help of a constant that is known as the equilibrium constant.
Complete step by step answer:
The enthalpy change of a reaction is determined as the total amount of heat absorbed or released when the reaction takes place. According to the given condition at equilibrium when ${N_2}{O_4}$ is heated, the following reaction will take place-
${N_2}{O_4} \rightleftharpoons 2N{O_2}$
According to the question, when the flask is heated to ${100^0}C$ the brown color deepens and on cooling, the brown colour becomes less coloured. So, according to le chatelier's principle when the temperature has raised the reaction will proceed in the forward direction and will favour the formation of more $N{O_2}$ compounds and due to this the solution becomes more brown. On decreasing the temperature, the equilibrium shift backward favors the reactant formation and the brown color will fade away. Based on these observations, we can say that these are the characteristics of an endothermic reaction. As we know, in endothermic reactions the energy of products is higher than the energy of reactants. So the change in enthalpy is given as:
$\Delta H = {\sum _{{{products}}}} - {\sum _{{{reactants}}}} = + ve$
Hence, we can say that the change in enthalpy $\Delta H$ of the system will be positive.
So, the correct answer is Option B.
Note: Endothermic reactions are those chemical reactions in which the reactants absorb heat energy from the surroundings to produce the products. In these reactions, the temperature of the surrounding area is lowered, which further creates a cooling effect.
Complete step by step answer:
The enthalpy change of a reaction is determined as the total amount of heat absorbed or released when the reaction takes place. According to the given condition at equilibrium when ${N_2}{O_4}$ is heated, the following reaction will take place-
${N_2}{O_4} \rightleftharpoons 2N{O_2}$
According to the question, when the flask is heated to ${100^0}C$ the brown color deepens and on cooling, the brown colour becomes less coloured. So, according to le chatelier's principle when the temperature has raised the reaction will proceed in the forward direction and will favour the formation of more $N{O_2}$ compounds and due to this the solution becomes more brown. On decreasing the temperature, the equilibrium shift backward favors the reactant formation and the brown color will fade away. Based on these observations, we can say that these are the characteristics of an endothermic reaction. As we know, in endothermic reactions the energy of products is higher than the energy of reactants. So the change in enthalpy is given as:
$\Delta H = {\sum _{{{products}}}} - {\sum _{{{reactants}}}} = + ve$
Hence, we can say that the change in enthalpy $\Delta H$ of the system will be positive.
So, the correct answer is Option B.
Note: Endothermic reactions are those chemical reactions in which the reactants absorb heat energy from the surroundings to produce the products. In these reactions, the temperature of the surrounding area is lowered, which further creates a cooling effect.
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