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Given the following data:

H2(g) + 1/2 O2(g) --> H20 (l) ∆H = -285.8 kJ

N2O5(g) + H20(l) --> 2HNO3(l) ∆H = -76.6 kJ

1/2 N2(g) + 3/2O2(g) + 1/2 H2 (g) --> HNO3(l) ∆ = -174.1 kJ

Calculate the ∆H for the reaction

2N2(g) + 5O2(g) --> 2N2O5(g)

**My Solution**

from N2: 2 = 1/2 *k3

from O2: 5 = 1/2 * k1 + 3/2 * k3

from N2O5: -2 = -k2

Hence ∆comb = ∆H1 * k1 + ∆H2 * k2 + ∆H3 * k3. I am not sure if this is right. Can someone please see if I made a mistake in my steps?

Thanks a lot