For the decomposition of calcium carbonate, consider the following thermodynamic data (due to variations in thermodynamic values for different sources, be sure to use the given values in calculating your answer.): δh∘rxn 178.5kj/molδs∘rxn 161.0j/(mol⋅k)calculate the temperature in kelvins above which this reaction is spontaneous. the standard free energy change, δg∘ , and the equilibrium constant k for a reaction can be related by the following equation: δg∘=−rtlnkwhere t is the kelvin temperature and r is equal to 8.314 j/(mol⋅k) .part bthe thermodynamic values from part a will be useful as you work through part b: δh∘rxn 178.5kj/molδs∘rxn 161.0j/(mol⋅k)calculate the equilibrium constant for the following reaction at room temperature, 25 ∘c : caco3(s)→cao(s)+co2(g)c)urea (nh2conh2) , an important nitrogen fertilizer, is produced industrially by the reaction2nh3(g)+co2(g)→nh2conh2(aq) +h2o(l)given that δg∘ = -13.6 kj , calculate δg at 25 ∘c for the following sets of conditions.40 atm nh3 , 40 atm co2 , 1.0 m nh2conh2 6.0×10−2 atm nh3 , 6.0×10−2 atm co2 , 1.0 m nh2conh2d)the first step in the commercial production of titanium metal is the reaction of rutile (tio2) with chlorine and graphite: tio2(s)+2cl2(g)+2c(s)→ticl4(l)+2co( g)use the following data to calculate δg∘ for this reaction at 25 ∘c .formula δh∘f(kj/mol) δg∘f(kj/mol) s∘(j/(k⋅mol))tio2(s) -944.0 -888.8 50.6cl2(g) 0 0 223.0c(s, graphite) 0 0 5.7ticl4(l) -804.2 -737.2 252.3co(g) -110.5 -137.2 197.6express your answer using one decimal place. calculate kp for this reaction at 25 ∘c .e)the standard free-energy change at 25 ∘c for the dissociation of water is 79.9 kj : 2h2o(l)⇌h3o+(aq)+oh−(aq)δg∘ = 79.9kjfor each of the following sets of concentrations, calculate δg at 25 ∘c and indicate whether the reaction is spontaneous in the forward or reverse direction.[h3o+] = [oh−] = 1.0 m[h3o+] = [oh−] = 1.0×10−7m[h3o+] = 8.0×10−8m,[oh−] = 8.0×10−11m use the thermodynamic data to calculate the equilibrium constant for the reaction. f)consider the haber synthesis of gaseous nh3 (δh∘f= −46.1kj/mol ; δg∘f=−16.5kj/mol) : n2(g)+3h2(g)→2nh3(g)use only these data to calculate δh∘ and δs∘ for the reaction at 25 ∘c .g)the molar solubility of lead iodide is 1.45×10−3m at 20 ∘c and 6.85×10−3m at 80 ∘c .what are the values of δh∘ and δs∘ for dissolution of pbi2 ? pbi2(s)→pb2+(aq)+2i−(aq)assume that δh∘ and δs∘ are independent of temperature.