The enthalpy change for the following reaction is -483.6 kJ:
2H2(g) + O2(g) = 2H2O(g)
Therefore, the enthalpy change for the following reaction is _______ kJ: 4H2(g) + 2O2(g) = 4H2O(g)
You have doubled the reaction; wouldn't the new enthalpy change be just twice as much.
To find the enthalpy change for the reaction 4H2(g) + 2O2(g) = 4H2O(g), we can use the fact that enthalpy change is a state function and is dependent on the stoichiometric coefficients of the balanced chemical equation.
Since the balanced equation shows that 2 moles of H2(g) react to form 2 moles of H2O(g) in the given reaction, we know that the enthalpy change for this reaction is the same as the given enthalpy change for the reaction 2H2(g) + O2(g) = 2H2O(g), which is -483.6 kJ.
Therefore, the enthalpy change for the reaction 4H2(g) + 2O2(g) = 4H2O(g) is also -483.6 kJ.
To find the enthalpy change for the second reaction, you can use the stoichiometry and enthalpy change of the first reaction.
The first reaction tells us that the enthalpy change for the reaction 2H2(g) + O2(g) = 2H2O(g) is -483.6 kJ.
Since the second reaction has twice the number of moles of hydrogen gas and oxygen gas, we can simply double the enthalpy change of the first reaction to get the enthalpy change for the second reaction.
Therefore, the enthalpy change for the reaction 4H2(g) + 2O2(g) = 4H2O(g) would be 2 * -483.6 kJ = -967.2 kJ.
Hence, the enthalpy change for the second reaction is -967.2 kJ.