Problem 5.4 Define points on 20C line with(geometry): point (p1 20, 44e6/(20+273),In(1e-4)): point(p220,53.2e6/(20+273),1n(1e-2)): get slope of line and multiply by 2R to get activation volume > Digits:=4:"V(m^3/mo1)'=2*8.314*s1ope(p220,p120); =.002440 ol Compute activation energy by horizontal difference between 20C and-60C lines >eq:=(V/2)*((69e6/(-60+273)) 44e6/(20+273)))=De1ta[H]*(1/(-60+273)-1/(20+273)); eq:=211.9= 62409 >'Delta[H] (kJ/mol)'=solve(eq, Delta[H])/1000;
Problem 5.4 Define points on 20C line > with(geometry):point(p1_20,44e6/(20+273),ln(1e-4)): > point(p2_20,53.2e6/(20+273),ln(1e-2)): get slope of line and multiply by 2R to get activation volume > Digits:=4:'V (m^3/mol)'= 2*8.314*slope(p2_20,p1_20); V 3 m mol = .002440 Compute activation energy by horizontal difference between 20C and -60C lines > eq:=(V/2)*((69e6/(-60+273)) - (44e6/(20+273)))=Delta[H]*(1/(-60+273) - 1/(20+273)); 80 eq := 211.9 = ∆H 62409 > 'Delta[H] (kJ/mol) ' = solve(eq,Delta[H])/1000; ∆H kJ mol = 165.3 Page 1
