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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]\[
C_2H_4 + 3O_2 \rightarrow 2CO_2 + 2H_2O, \Delta H = -1410 \text{ kJ}
\][/tex]

A. [tex]\(-1410 \text{ kJ}\)[/tex]
B. [tex]\(1410 \text{ kJ}\)[/tex]
C. [tex]\(-2820 \text{ kJ}\)[/tex]
D. [tex]\(2820 \text{ kJ}\)[/tex]

Sagot :

GinnyAnsweridn
To reverse the given reaction, we need to change the direction of the reaction, which in turn will change the sign of the enthalpy change (ΔH). The given reaction is:

[tex]\[ \ce{C2H4 + 3 O2 -> 2 CO2 + 2 H2O} \quad (\Delta H = -1410 \text{ kJ}) \][/tex]

When we reverse this reaction, the reactants become products and the products become reactants. The reversed reaction is:

[tex]\[ \ce{2 CO2 + 2 H2O -> C2H4 + 3 O2} \][/tex]

For the enthalpy change associated with the reverse reaction, the sign of ΔH will change. Therefore, the enthalpy change for the reversed reaction becomes:

[tex]\[ \Delta H = +1410 \text{ kJ} \][/tex]

Thus, the final value for the enthalpy of reaction you use for this reversed intermediate reaction is:

[tex]\[ \boxed{1410 \text{ kJ}} \][/tex]

So, the correct answer is B. 1410 kJ
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