Post Lecture Homework Chapter 06 nteractive Activity-Applications of Hess\'s Law

Question

Post Lecture Homework Chapter 06 nteractive Activity-Applications of Hess's Law 10 of 13> Part B Changes sign when a Click on the the two reactions that are displayed. The reaction that was on the screen when you started and its derivative demonstrate that the reaction enthalpy, A, changes sign when a process is reversed is revernedbutton within the activity and analyze the relationship between Consider the reaction What will Arbe for the reaction if it is reversed? Express your answer with appropriate units View Available Hint(s) Value Units Submit Hess's law of constant heat summation Hess's law states that if a reaction is carried out in a series of steps, AH for the overall reaction will equal the sum of the enthalpy changes for the individual steps. By using Hess's law, you can calculate the enthalpy of the reaction. For example, you can calculate the reaction enthalpy for the combustion of solid sulfur, s(s), to form so, gas The combustion of solid sulfur to form so, gas occurs in the two steps Step 1 : S(s) +02(g)SO2(g),aH =-297 kJ Step 2: SO2(g)- 02(g)-SO3(g), »--99 kJ The overall reaction will be obtained by adding these two steps. When you add the two steps, the reaction enthalpy values for the individual steps should also be added. Remember that when you add the two steps, the molecule that appears on both sides of a reaction arrow is cancelled from both sides of the equation Thus, by adding step 1 and step 2, you can obtain the reaction enthalpy for the overall reaction H =-297 kJ S(s) + 02 (g) SO3(g) H=-396 kJ Hence, the reaction enthalpy for the overall reaction is-297 k-99kJ)396 kJ Part C

Explanation / Answer

Ans :

Part B :

When the reaction is reversed , the sign for the enthalpy change reverses , but the value remains same .

So here : -(-1270 KJ)

= + 1270 KJ

Part C :

The desired reaction can be obtained by adding :

= (reverse of equation 1 ) + ( equation 2 ) + (equation 3) + (reverse of equation 4 multiplied by 2)

So the enthalpy change :

= (+74.8) + (-393.5) + (-484.0) + 2(-44.0)

= -890.7 KJ

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