Question

Given the information below, calculate ∆?∘?,298.15? and ∆?∘?,400? for the following reaction:

16. Given the information below, calculate ΔΗο,298.15K and AHR,400K for the following reaction: Mg(g) + 2HCl(g) → H2(g) + MgCl2(s) Assume that heat capacities are constant over the desired temperature range. All molar enthalpies of formation are at 298.15K. AH; Mg(g) = 147.1 kj/mol CP,mMg(g) = 20.8 J mol 'K-1 Gun HCIG) = 29.1 J mol K! G,in H2(g) = 28.81 mol: K- CP,m MgCl2(s) = 71.4 J moll K" AHİ MgCl2(s) =-641.3 kj/mol SEP 18 認 2A MacBook Air

??(?) + 2???(?) → ?2(?) + ????2(?)

Assume that heat capacities are constant over the desired temperature range. All molar enthalpies of formation are at 298.15K.

∆?∘ Mg(g) ? = 147.1 kJ/mol

∆?∘ HCl(g) ? = -92.3 kJ/mol

∆?∘ H2(g) ? = 0 kJ/mol

∆?∘ MgCl2(s) = -641.3 kJ/mol

Cp,m Mg(g) = 20.8 J mol-1 K-1
Cp,m HCl(g) = 29.1 J mol-1 K-1
Cp,m H2(g) = 28.8 J mol-1 K-1

Cp,m MgCl2(s) = 71.4 J mol-1 K-1

Cp,m MgCl2(s) = 71.4 J mol-1 K-1

Answer 1

Reaction: Mg (g) + 2 HCl (g) rightarrow H_2 (g) + Mg Cl_2 (s) angle H = ? kJ Standard enthalpies of the formation of the substances involved in the reaction: Standard enthalpy change of the reaction is, angle H^0_1 times n = m times angle H^0_f (products) - n times angle H^0_f (reactants) = [(1 times angle H^0_f (H_2) + 1 times angle H^0_f (Mg Cl_2)) - (1 times angle H^0_f (Mg) + 2 times angle H^0_f (HCl))] = [((1 mol times 0 kJ/mol + 1 mol times -641.3 kJ/mol)) - ((1 mol times 147.1 kJ/mol) + (2 mol times -92.3 kJ/mol))] = -603.8 kJ Therefore, the standard enthalpy of reaction is -603.8 kJ. angle C_p = mC_p (products) - nC_p (reactants) = (1 mol times C_p H_2 (g) + 1 mol times C_p Mg Cl_2 (s)) - (1 mol times C_p Mg (s) + 2 mol times C_p HCl (g)) = (1 times 28.8