Question

# Given the following data: HNO3(l) → 1/2N2(g) + 3/2O2(g) + 1/2H2(g) ΔH° = 174.1 kJ 2N2(g)...

Given the following data:

 HNO3(l) → 1/2N2(g) + 3/2O2(g) + 1/2H2(g) ΔH° = 174.1 kJ 2N2(g) + 5O2(g) → 2N2O5(g) ΔH° = 28.4 kJ H2(g) + 1/2O2(g) → H2O(l) ΔH° = -285.8 kJ

Calculate ΔH° for the reaction:
2HNO3(l) → N2O5(g) + H2O(l)
Note that you should be able to answer this one without needing to use any additional information from the thermo table.

HNO3(l) → 1/2N2(g) + 3/2O2(g) + 1/2H2(g)   : ΔH°1 = 174.1 kJ     -------(1)

2N2(g) + 5O2(g) → 2N2O5(g)    ; ΔH°2 = 28.4 kJ                        -------(2)

H2(g) + 1/2O2(g) → H2O(l)     ; ΔH°3 = -285.8 kJ                        ------(3)

2HNO3(l) → N2O5(g) + H2O(l)   : ΔH° = ?                               -------(4)

Eqn (4) can be obtained from remaining equations as follows:

Eqn(1) = [2xEqn(1)] + [(1/2)xEqn(2)] +Eqn(3)

So ΔH° = [2x ΔH°1 ] + [ (1/2)x ΔH°2 ] + ΔH°3

= [2x(174.1 kJ)] + [(1/2)x(28.4 kJ)] + (-285.8 kJ)

= +76.6 kJ

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