the decomposition of calcium carbonate, caco3(s) → cao(s) + co2(g), has the following values for free energy…

the decomposition of calcium carbonate, caco3(s) → cao(s) + co2(g), has the following values for free energy and enthalpy at 25.0°c. δg = 130.5 kj/mol δh = 178.3 kj/mol what is the entropy of the reaction? use δg = δh - tδs. -160.3 j/(mol k) -47.8 j/(mol k) 160.3 j/(mol k) 1,912 j/(mol k)

the decomposition of calcium carbonate, caco3(s) → cao(s) + co2(g), has the following values for free energy and enthalpy at 25.0°c. δg = 130.5 kj/mol δh = 178.3 kj/mol what is the entropy of the reaction? use δg = δh - tδs. -160.3 j/(mol k) -47.8 j/(mol k) 160.3 j/(mol k) 1,912 j/(mol k)

Answer

Answer:

-47.8 J/(mol K)

Explanation:

Step1: Convert temperatura a Kelvin

$T = 25.0 + 273.15=298.15\ K$

Step2: Reorganizar la ecuación $\Delta G=\Delta H - T\Delta S$

$\Delta S=\frac{\Delta H-\Delta G}{T}$

Step3: Sustituir valores

$\Delta S=\frac{178.3\times10^{3}\ J/mol - 130.5\times10^{3}\ J/mol}{298.15\ K}$ $\Delta S=\frac{47.8\times10^{3}\ J/mol}{298.15\ K}\approx - 47.8\ J/(mol\ K)$