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Multiple Choice
The enthalpy change for the following reaction is -483.6 kJ: 2 H2 (g) + O2 (g) --> 2 H2O (g). Therefore, the enthalpy change for the following reaction is ________ kJ: 6 H2O (g) --> 6 H2 (g) + 3 O2 (g).
A
-1450.8 kJ
B
1450.8 kJ
C
-483.6 kJ
D
483.6 kJ
Verified step by step guidance
1
Identify the given reaction and its enthalpy change: 2 H2 (g) + O2 (g) --> 2 H2O (g) with ΔH = -483.6 kJ.
Recognize that the enthalpy change is for the formation of 2 moles of H2O (g).
The problem asks for the enthalpy change of the reverse reaction: 6 H2O (g) --> 6 H2 (g) + 3 O2 (g).
Reverse the original reaction: 2 H2O (g) --> 2 H2 (g) + O2 (g). The enthalpy change for the reverse reaction is the opposite sign, so ΔH = +483.6 kJ for 2 moles of H2O.
Scale the reaction to match the desired reaction: Multiply the reverse reaction by 3 to get 6 H2O (g) --> 6 H2 (g) + 3 O2 (g). Multiply the enthalpy change by 3 as well: ΔH = 3 * 483.6 kJ.