Respuesta :
For the first question, the correct answer among the choices listed is option A, a chocolate bar melting in hot car. It is an endothermic process because it absorbs heat energy for it to be melted. For the second question, it think the correct answer is the last option, (-2222) + (-1182) - (-1144).
Answer:
For 1: The correct answer is Option a.
For 2: The correct answer is (-2222) - [(-1182) + (-1144)] kJ
Explanation:
- For 1:
Endothermic reactions are defined as the reactions in which energy is absorbed by the system.
Exothermic reactions are defined as the reactions in which heat is released by the system.
For the given options:
Option a: In this case, solid chocolate bar is meting with increase in temperature. Thus, it is absorbing heat from the surrounding and is considered as an endothermic reaction.
Option b: In this case, fruit are freezing in the freezer. Thus, it is releasing heat to the surrounding and is considered as an exothermic reaction.
Option c: Combustion reactions are defined as the reactions in which hydrocarbon reacts with oxygen gas to release carbon dioxide and water molecule with the release of heat. Thus, it considered as an exothermic reaction.
Option d: In this case, water molecules present in the air are getting condensed. Thus, it is releasing heat to the surrounding and is considered as an exothermic reaction.
From the above information, the correct answer is option a.
- For 2:
According to Hess’s law, the heat absorbed or evolved in chemical equation is same whether the process is occuring in one step or in several steps.
According to this law, the chemical equation is treated as the ordinary algebraic expression which can be added or subtracted to get the required equation. This means the enthalpy change of the overall reaction will be equal to the sum of the enthalpy changes of the intermediate reactions.
We are given 3 intermediate balanced reactions:
Equation 1: [tex]3C+3O_2\rightarrow 3CO_2;\Delta H_1=-1182kJ[/tex]
Equation 2: [tex]4H_2+2O_2\rightarrow 4H_2O;\Delta H_2=-1144kJ[/tex]
Equation 3: [tex]C_3H_8+5O_2\rightarrow 3CO_2+4H_2O;\Delta H_3=-2222kJ[/tex]
Final Equation: [tex]C_3H_8\rightarrow 3C+4H_2[/tex]
By adding Equation 3 and reverse reaction of Equation 1 and Equation 2, we will get the enthalpy of final equation.
[tex]\Delta H_{final}=\Delta H_{3}-[\Delta H_1+\Delta H_2]\\\\\Delta H_{final}=-2222+[(-1182)+(-1144)]kJ[/tex]
Hence, the correct answer is (-2222) - [(-1182) + (-1144)] kJ