Question

Calculate ∆G° for the equation below
using the thermodynamic data given
N ² + 3 H ²➡️ 2 NH³(g)

∆Hº (NH³) = -46 kJ; ∆ S' values (J/mol-K)
are
NH³ = 192.5, N² = 191.5, H² = 130.6.
1. +33 kJ
2. +66 kJ
3. -66 kJ
4. -33 kJ​

Answer 1

Answer:

$30 \frac{2 = \div \sqrt[31 = 2 { = .13 \frac{225.}{?} }^{2} \times \frac{?}{?} ]{?} }{?} = . = .3.516842 {5456}^{?} 22255y22 = = \times \frac{?}{?} 5792)6663x$