Tes Online Materi Termokimia

1. Pernyataan yang benar mengenai reaksi eksoterm adalah…

Berlangsung tidak spontan

Perubahan memerlukan kalor

Perubahan entalpi positif

Terjadi perpindahan kalor dari sistem ke lingkungan

The correct answer states that there is a transfer of heat from the system to the surroundings. This is a characteristic of an exothermic reaction, where energy is released in the form of heat. It is opposite to an endothermic reaction, where heat is absorbed from the surroundings. The other statements are incorrect because an exothermic reaction can still be spontaneous, the change in enthalpy can be negative, and the reaction can still require a certain amount of heat energy.

Explanation

The correct answer states that there is a transfer of heat from the system to the surroundings. This is a characteristic of an exothermic reaction, where energy is released in the form of heat. It is opposite to an endothermic reaction, where heat is absorbed from the surroundings. The other statements are incorrect because an exothermic reaction can still be spontaneous, the change in enthalpy can be negative, and the reaction can still require a certain amount of heat energy.

2. Kapur tulis dilarutkan ke dalam tabung reaksi berisi air. Tabung reaksi mengalami kenaikan suhu saat reaksi terjadi. Reaksi tersebut termasuk …

eksoterm

endoterm

pembakaran

oksidasi

The correct answer is "eksoterm." This is because the question states that the reaction causes an increase in temperature in the reaction tube. In an exothermic reaction, heat is released to the surroundings, resulting in a temperature increase. Therefore, the given information aligns with the characteristics of an exothermic reaction.

Explanation

The correct answer is "eksoterm." This is because the question states that the reaction causes an increase in temperature in the reaction tube. In an exothermic reaction, heat is released to the surroundings, resulting in a temperature increase. Therefore, the given information aligns with the characteristics of an exothermic reaction.

3. Logam natrium dilarutkan dalam tabung reaksi berisi air. Tabung reaksi mengalami kenaikan suhu saat reaksi terjadi. Sistem pada reaksi di atas meliputi …

Tabung reaksi

Logam natrium dan air

Udara

Logam natrium, air, dan tabung reaksi

The correct answer is "Logam natrium dan air" because the question states that when sodium metal is dissolved in water, there is an increase in temperature. This indicates that a chemical reaction is taking place between the sodium metal and water, resulting in the release of energy in the form of heat. Therefore, the system involved in this reaction includes only the sodium metal and water. The presence of the reaction tube and air is not directly related to the increase in temperature.

Explanation

The correct answer is "Logam natrium dan air" because the question states that when sodium metal is dissolved in water, there is an increase in temperature. This indicates that a chemical reaction is taking place between the sodium metal and water, resulting in the release of energy in the form of heat. Therefore, the system involved in this reaction includes only the sodium metal and water. The presence of the reaction tube and air is not directly related to the increase in temperature.

4. Reaksi pembakaran CH4 menghasilkan kalor sebesar 802 kJ/mol. Pada pembakaran 3,2 gram CH4 dibebaskan kalor sebesar … kJ.(Ar C= 12, H= 1)

80,2

120,3

160,4

240,6

The given question asks for the amount of heat released when 3.2 grams of CH4 is burned. To find the heat released, we need to use the molar mass of CH4 and the given heat of combustion. The molar mass of CH4 is calculated by adding the atomic masses of carbon and hydrogen, which gives us 16 grams/mol. Since we have 3.2 grams of CH4, we can divide it by the molar mass to find the number of moles, which is 0.2 mol. Finally, we can multiply the number of moles by the given heat of combustion (802 kJ/mol) to find the heat released, which is 160.4 kJ.

Explanation

The given question asks for the amount of heat released when 3.2 grams of CH4 is burned. To find the heat released, we need to use the molar mass of CH4 and the given heat of combustion. The molar mass of CH4 is calculated by adding the atomic masses of carbon and hydrogen, which gives us 16 grams/mol. Since we have 3.2 grams of CH4, we can divide it by the molar mass to find the number of moles, which is 0.2 mol. Finally, we can multiply the number of moles by the given heat of combustion (802 kJ/mol) to find the heat released, which is 160.4 kJ.

5. Diketahui reaksi : NO(g) à ½ N2(g) + ½ O2(g)Jika entalpi N2 dan O2 lebih besar dari entalpi NO, maka reaksi tersebut merupakan reaksi …

Penguraian

Pembakaran

endoterm

eksoterm

The given reaction is the decomposition of NO into N2 and O2. If the enthalpy of N2 and O2 is greater than the enthalpy of NO, it means that more energy is required to break the bonds in NO and form N2 and O2. This indicates that the reaction is endothermic, as it requires an input of energy to proceed.

Explanation

The given reaction is the decomposition of NO into N2 and O2. If the enthalpy of N2 and O2 is greater than the enthalpy of NO, it means that more energy is required to break the bonds in NO and form N2 and O2. This indicates that the reaction is endothermic, as it requires an input of energy to proceed.

6. Entalpi pembentukan CO2, H2O, dan C2H6 berturut-turut -394 kJ, -286 kJ, dan -76 kJ. Entalpi pembakaran C2H6 sebesar … kJ.

-920

-870

-1570

-270

The enthalpy of formation for CO2, H2O, and C2H6 are given as -394 kJ, -286 kJ, and -76 kJ respectively. The enthalpy of combustion for C2H6 can be calculated by subtracting the sum of the enthalpies of formation of the products (CO2 and H2O) from the sum of the enthalpies of formation of the reactants (C2H6). Therefore, -394 kJ + (-286 kJ) = -680 kJ is the enthalpy change for the formation of CO2 and H2O. Subtracting this from the enthalpy of formation of C2H6 (-76 kJ) gives us -680 kJ - (-76 kJ) = -604 kJ as the enthalpy of combustion for C2H6. However, none of the given answer choices match this calculation, so the correct answer is not available.

Explanation

The enthalpy of formation for CO2, H2O, and C2H6 are given as -394 kJ, -286 kJ, and -76 kJ respectively. The enthalpy of combustion for C2H6 can be calculated by subtracting the sum of the enthalpies of formation of the products (CO2 and H2O) from the sum of the enthalpies of formation of the reactants (C2H6). Therefore, -394 kJ + (-286 kJ) = -680 kJ is the enthalpy change for the formation of CO2 and H2O. Subtracting this from the enthalpy of formation of C2H6 (-76 kJ) gives us -680 kJ - (-76 kJ) = -604 kJ as the enthalpy of combustion for C2H6. However, none of the given answer choices match this calculation, so the correct answer is not available.

7. Diketahui reaksi : SO2(g) + ½ O2(g) à SO3(g) Memiliki ∆H = -2,5 kJ.Nilai ∆H tersebut merupakan …

Entalpi pembakaran SO2

Entalpi penguraian SO3

Entalpi pembakaran SO3

Entalpi pembentukan SO2

The given reaction is the combustion of SO2, which means that SO2 is reacting with O2 to form SO3. The negative value of ∆H (-2.5 kJ) indicates that the reaction is exothermic, releasing heat. Therefore, the correct answer is "Entalpi pembakaran SO2" which translates to "Enthalpy of combustion of SO2".

Explanation

The given reaction is the combustion of SO2, which means that SO2 is reacting with O2 to form SO3. The negative value of ∆H (-2.5 kJ) indicates that the reaction is exothermic, releasing heat. Therefore, the correct answer is "Entalpi pembakaran SO2" which translates to "Enthalpy of combustion of SO2".

8. Diketahui persamaan reaksi termokimia sebagai berikut.2C + H2 à C2H2 ∆H = -a kJ/molC + O2 à CO2 ∆H = -b kJ/molH2 + ½ O2 à H2O ∆H = -c kJ/molBanyaknya kalor pembakaran 1 mol gas asetilen menurut Hukum Hess sebesar … kJ.

(a-2b-c)

(a+ 2b + c)

(-a-2b-c)

(-a-2b+c)

The given question provides the thermochemical equations for the combustion of acetylene and the formation of carbon dioxide and water. According to Hess's Law, the overall enthalpy change for a reaction can be calculated by summing the enthalpy changes of the individual reactions involved. In this case, the enthalpy change for the combustion of 1 mol of acetylene can be determined by subtracting the enthalpy changes of the formation of carbon dioxide and water from the enthalpy change of the formation of acetylene. Therefore, the correct answer is (a-2b-c).

Explanation

The given question provides the thermochemical equations for the combustion of acetylene and the formation of carbon dioxide and water. According to Hess's Law, the overall enthalpy change for a reaction can be calculated by summing the enthalpy changes of the individual reactions involved. In this case, the enthalpy change for the combustion of 1 mol of acetylene can be determined by subtracting the enthalpy changes of the formation of carbon dioxide and water from the enthalpy change of the formation of acetylene. Therefore, the correct answer is (a-2b-c).

9. Diantara reaksi berikut yang termasuk reaksi endoterm adalah…

2 HBr(g)  H2(g) + Br2(g) +a kkal

CaO(s)  Ca(s) +1/2 O2(g) –b kkal

Fe2O3(g) + 4H2(g)  3Fe(s) + 4H2O(g) –c kkal

Ca(s) + C(s) + 3/2 O2(g)  CaCO3(s) + e kkal

The given reaction CaO(s) → Ca(s) + 1/2 O2(g) is an example of an endothermic reaction because it requires energy to break the bonds in the reactants and form the products. In this reaction, calcium oxide (CaO) is being decomposed into calcium (Ca) and oxygen gas (O2). Since the reaction is absorbing energy (represented by -b kkal), it is classified as endothermic.

Explanation

The given reaction CaO(s) → Ca(s) + 1/2 O2(g) is an example of an endothermic reaction because it requires energy to break the bonds in the reactants and form the products. In this reaction, calcium oxide (CaO) is being decomposed into calcium (Ca) and oxygen gas (O2). Since the reaction is absorbing energy (represented by -b kkal), it is classified as endothermic.

10. Diketahui :∆Hof H2O(l) = -78,3 kkal∆Hof CO2(g) = -94 kkal∆Hof C3H7COOH(l) = -78,3 kkal∆H reaksi pembakaran C3H7COOH sebesar … kkal.

+1.135,6

+612,2

+564,3

-564,3

The enthalpy change of combustion can be calculated by subtracting the sum of the enthalpies of the reactants from the sum of the enthalpies of the products. In this case, the reactants are C3H7COOH(l) and O2(g), and the products are CO2(g) and H2O(l). The enthalpy change of combustion can be expressed as:

∆H combustion = ∆Hof CO2(g) + ∆Hof H2O(l) - ∆Hof C3H7COOH(l)

Substituting the given values:

∆H combustion = -94 kkal + (-78,3 kkal) - (-78,3 kkal)
∆H combustion = -94 kkal + 78,3 kkal + 78,3 kkal
∆H combustion = 62,2 kkal

Therefore, the correct answer is +612,2.

Explanation

The enthalpy change of combustion can be calculated by subtracting the sum of the enthalpies of the reactants from the sum of the enthalpies of the products. In this case, the reactants are C3H7COOH(l) and O2(g), and the products are CO2(g) and H2O(l). The enthalpy change of combustion can be expressed as:

∆H combustion = ∆Hof CO2(g) + ∆Hof H2O(l) - ∆Hof C3H7COOH(l)

Substituting the given values:

∆H combustion = -94 kkal + (-78,3 kkal) - (-78,3 kkal)
∆H combustion = -94 kkal + 78,3 kkal + 78,3 kkal
∆H combustion = 62,2 kkal

Therefore, the correct answer is +612,2.