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Chemistry Question on Thermodynamics

For the reaction, CaCO3(g)<=>CaO(s)+CO2(g) {CaCO_{3(g)}<=>CaO_{(s)} +CO_{2(g)}} partial pressure of CO2CO_2 at 1000 K1000\, K is 0.0030.003 atm. ?G?=27.2 kcal?G^? = 27.2\, kcal. Calculate the value of ?G.?G.

A

12.6 kcal12.6\, kcal

B

15.6 kcal15.6\, kcal

C

13.4 kcal13.4\, kcal

D

14.2 kcal14.2\, kcal

Answer

15.6 kcal15.6\, kcal

Explanation

Solution

CaCO3(g)<=>CaO(s)+CO2(g) {CaCO_{3(g)}<=>CaO_{(s)} +CO_{2(g)}} Kp=pCO2=0.003K_p =p_CO_2 =0.003 atm ?G=?G?+2.303 RT logKp?G = ?G^? + 2.303\,RT\, logK_p =27200+2.303×2×1000×log 0.003= 27200 + 2.303 \times 2 \times 1000 \times log \,0.003 =15.6 kcal.(R=2 cal mol−1 K−1)= 15.6\, kcal. (R = 2\, cal\, mol^{-1}\, K^{-1})