What is the change in enthalpy for the following reaction?
4nh3(g) + 302(g) 2n2(g) + 6h2o(l)
given:
nh3: ∆h= -46.2 kj
given: h2o: ∆h= -286 kj
a. -1530.4 kj
b. -1020.3 kj
c. 1530.4 kj
d. 1020.3 kj
e. -510.1 kj

A. -1530.4 kJ

Since; NH3: ∆H= -46.2 kJ

          H2O: ∆H= -286 kJ

Then; ΔH = 4(-46.2) + 3(-286)

      = -1530.4 kJ

This means the reaction is exothermic as ΔH is negative.

the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be