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authorprashantsinalkar2017-10-10 12:27:19 +0530
committerprashantsinalkar2017-10-10 12:27:19 +0530
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+//ques-18.28
+//Calculating heat of reaction in terms of calories
+clc
+K1=1.64*10^-4; K2=0.144*10^-4;//equilibrium constant (in atm)
+T1=273+400; T2=273+500;//temperature (in K)
+R=1.987;//cal/mol/K
+H=(log10(K2/K1)*2.303*R*T1*T2)/(T2-T1);
+printf("The heat of the reaction is %.2f kcal/mol.",H/1000);