If you need to reverse the following reaction in order for it to be an intermediate reaction in a Hess's law problem, what would be the final value for the enthalpy of reaction you use for this intermediate reaction?
[tex]C_2 H_4+3 O_2 \rightarrow 2 CO_2+2 H_2 O, \Delta H=-1410 kJ[/tex]
A. 2820 kJ
B. -2820 kJ
C. 1410 kJ
D. -1410 kJ
Answer (2)
Reverse the reaction: 2 C O 2 + 2 H 2 O → C 2 H 4 + 3 O 2 .
Change the sign of Δ H : Δ H re v erse = − ( Δ H f or w a r d ) .
Calculate the new Δ H : Δ H re v erse = − ( − 1410 kJ ) = 1410 kJ .
The final enthalpy of reaction is 1410 kJ .
Explanation
Understanding the Problem We are given a chemical reaction and its enthalpy change, Δ H . The reaction is: C 2 H 4 + 3 O 2 → 2 C O 2 + 2 H 2 O , Δ H = − 1410 kJ We need to find the enthalpy change for the reverse reaction.
Reversing the Reaction When a reaction is reversed, the sign of the enthalpy change, Δ H , is also reversed. This is because the energy that was released in the forward reaction must now be supplied to drive the reverse reaction.
Calculating the New Enthalpy The reverse reaction is: 2 C O 2 + 2 H 2 O → C 2 H 4 + 3 O 2 The enthalpy change for the reverse reaction is the negative of the enthalpy change for the forward reaction: Δ H reverse = − ( Δ H forward ) = − ( − 1410 kJ ) = 1410 kJ Therefore, the enthalpy change for the reversed reaction is 1410 kJ.
Examples
Reversing a chemical reaction is similar to understanding how energy changes when you melt ice versus when water freezes. Melting ice requires energy (endothermic, positive Δ H ), while freezing water releases energy (exothermic, negative Δ H ). If melting ice needs 334 J/g, then freezing water releases 334 J/g. Similarly, in chemical reactions, reversing the process simply changes the sign of the energy involved.
To find the enthalpy change for the reversed reaction of the given combustion reaction, we reverse its sign. The final value for this intermediate reaction's enthalpy is 1410 kJ. Therefore, the correct answer is option C: 1410 kJ.
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