The standard internal energy change for a reaction can be symbolized as δu°rxn or δe°rxn. for the following reaction equations, calculate the energy change of the reaction at 25 °c and 1.00 bar. sn(s)+cl2(g) > sncl4(l) + h(kjmol) -511.3. h2(g)+cl2(g) > 2hcl(g) + h(kjmol) -184.6

The formula for ΔU°rxn is: ΔU°rxn = H - PV. Assuming ideal gas, it could also be: ΔU°rxn = H - RT.
 For the first reaction:
 ΔU°rxn = (-511.3 kJ/mol) - (0.008314 J/mol-K)(25+273 K) = -513.78 kJ/mol
 For the second reaction:
ΔU°rxn = (-184.6 kJ/mol) - (0.008314 J/mol-K)(25+273 K) = -187.08 kJ/mol

Answerthis solution will not form a precipitateexplanationsolubility equilbrium:         pbf₂(s)  ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is    (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as    then, we can interpret  q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually  forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product:     since q = 2.7  × 10⁻⁸ < ksp =  4  ×  10⁻⁸, a precipitate will not form.

Answer the air near the ground cools to its dew point