Given the standard enthalpy changes for the following two reactions:
(1) N2+2O2 -> N2O4 Δ H=9.2kj
(2) 2N2O -> 2N2+O2 Δ H=-164.2kj
What is the standard enthalpy change for the following reaction?
(3) 2N2O+3O22N2O4 Δ H=?
Standard enthalpy change = ________ kJ
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To find the standard enthalpy change for reaction (3), we can combine reactions (1) and (2).
First, we need to reverse reaction (1) to obtain N2O4 as a product instead of a reactant. Since reversing the reaction changes the sign of the enthalpy change, we now have:
(1) N2O4 -> N2+2O2 Δ H=-9.2 kj
Next, we need to multiply reaction (2) by 2 to obtain 2N2O as a reactant. Multiplying the reaction also multiplies the enthalpy change, so we now have:
(2) 4N2O -> 4N2+2O2 Δ H=-328.4 kj
Now we can add the two reactions together to obtain reaction (3):
(3) 2N2O+3O2 -> 2N2O4
Adding the enthalpy changes, we get:
ΔH = (-9.2 kj) + (-328.4 kj) = -337.6 kj
Thus, the standard enthalpy change for reaction (3) is -337.6 kJ.
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