Classify the four reactions that your group analyzed today as being either endothermic or exothermic

1 answer:

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Chemical reactions are basically divided into two major classes depending on whether the reaction lose energy or gain energy from the environment during the course of the reaction. The two classes of reaction are exothermic and endothermic reaction.
An exothermic reaction is a type of reaction in which the reaction system lose energy to the environment and thus, the energy content of the reactants is more than that of the product formed. Because of this, the enthapyl change of an exothermic reaction is always negative.
An endothermic reaction is a type of reaction in which the reaction system absorb energy from the environment. Thus, the energy contents of the products is always higher than that of the reactants and the enthapyl change of the reaction is always positive. During the course of the reaction, the reaction container is usually cold to the touch because energy is been absorbed from the environment.

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Consider the intermediate chemical reactions. 2 equations. First: upper C a (s) plus upper C upper O subscript 2 (g) plus one ha

DochEvi [55]

<u>Answer:</u> When the enthalpy of this overall chemical equation is calculated, the enthalpy of the second intermediate equation is halved and has its sign changed.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The overall chemical reaction follows:

$CaO(s)+CO_2\rightarrow CaCO_3(s)$ $\Delta H^o_{rxn}=?$

The intermediate balanced chemical reaction are:

(1) $Ca(s)+CO_2(g)+\frac{1}{2}O_2(g)\rightarrow CaCO_3(s)$ $\Delta H_1=-812.8kJ$

(2) $2Ca(s)+O_2(g)\rightarrow 2CaO(s)$ $\Delta H_2=-1269kJ$

The expression for enthalpy of the reaction follows:

$\Delta H^o_{rxn}=[1\times (\Delta H_1)]+[\frac{1}{2}\times (-\Delta H_2)]$

Hence, when the enthalpy of this overall chemical equation is calculated, the enthalpy of the second intermediate equation is halved and has its sign changed.

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Answer:

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larisa86 [58]

Tin would be the right answer

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