1. Find the equilibrium constant of the reaction: NH_(3)(g)+HCl(g)=NH_(4)Cl(s) at 298K if G_(298)(NH_(3)(g))=-16.7kJ/mol,C_(298)(HCl(g))=-95.4kJ/mol and G_(298)(NH_(4)Cl(s))=-203.7kJ/mol

To find the equilibrium constant (K) for the reaction at 298K, we can use the Gibbs free energy change (ΔG) and the equation:<br /><br />ΔG = -RT ln K<br /><br />where R is the gas constant (8.314 J/mol·K) and T is the temperature in Kelvin.<br /><br />First, we need to calculate the standard Gibbs free energy change (ΔG°) for the reaction:<br /><br />ΔG° = G°(products) - G°(reactants)<br />ΔG° = G°(NH4Cl(s)) - [G°(NH3(g)) + G°(HCl(g))]<br />ΔG° = -203.7 kJ/mol - (-16.7 kJ/mol - 95.4 kJ/mol)<br />ΔG° = -203.7 kJ/mol + 112.1 kJ/mol<br />ΔG° = -91.6 kJ/mol<br /><br />Now, we can calculate the equilibrium constant (K) using the equation:<br /><br />ΔG° = -RT ln K<br />-91.6 kJ/mol = -(8.314 J/mol·K)(298 K) ln K<br />ln K = -91.6 kJ/mol / (-8.314 J/mol·K)(298 K)<br />ln K = 38.4<br />K = e^38.4<br /><br />Since this value is extremely large, it indicates that the reaction strongly favors the formation of NH4Cl(s) under the given conditions.