Question

Calculate the standard enthalpy change for the formation of ammonia gas from nitrogen gas and hydrogen gas using Hess's Law, given the following equations and their corresponding enthalpy changes:N2(g) + 3H2(g) → 2NH3(g)   ∆H = -92.4 kJ/molN2(g) → 2N(g)                ∆H = 941 kJ/molH2(g) → 2H(g)                 ∆H = 436 kJ/mol

Answer 1

To use Hess's Law, we need to manipulate the given equations in such a way that when they are added together, they result in the target equation: N2 g  + 3H2 g   2NH3 g . First, let's look at the second and third equations. We need to reverse them so that N and H atoms are on the product side:1. 2N g   N2 g                 H = -941 kJ/mol2. 2H g   H2 g                 H = -436 kJ/molNow, we need to multiply the second equation by 3 to account for the 3 moles of hydrogen gas in the target equation:2H g   H2 g                 H = -436 kJ/molx 36H g   3H2 g               H = -1308 kJ/molNow, we can add the first and modified second and third equations together:1. 2N g   N2 g                 H = -941 kJ/mol2. 6H g   3H2 g               H = -1308 kJ/mol----------------------------------------------   2N g  + 6H g   N2 g  + 3H2 g   H = -2249 kJ/molFinally, we can use the target equation's enthalpy change to find the enthalpy change for the formation of ammonia:N2 g  + 3H2 g   2NH3 g    H = -92.4 kJ/mol2N g  + 6H g   N2 g  + 3H2 g   H = -2249 kJ/molN2 g  + 3H2 g   2NH3 g    H = -92.4 kJ/mol------------------------------------------------2N g  + 6H g   2NH3 g     H = -2156.6 kJ/molSince the target equation has 2 moles of NH3, we need to divide the enthalpy change by 2 to find the enthalpy change for the formation of 1 mole of ammonia:H = -2156.6 kJ/mol  2 = -1078.3 kJ/molSo, the standard enthalpy change for the formation of ammonia gas from nitrogen gas and hydrogen gas is -1078.3 kJ/mol.