oo AT&T; 11:00 PM * 32%. psu.instructure.com DQuestion 6 1 pts Thermodynamic Qua

Question

oo AT&T; 11:00 PM * 32%. psu.instructure.com DQuestion 6 1 pts Thermodynamic Quantities for Selected Substances at 298.15 K (25°C (kJ/mol mol) NOCKg)52.6 66.3 264 NOg)90.37 6.71210.62 Cls, 0 0 5.69 CCLg)106.7-64.0 309.4 SO2g)296.9300.4 248.5 S(s, 0 0 31.88 MgCl2ls)6416 591 89.6 MgOs)60185696 26.8 H20 (g)24182 -228.57 188.83 H20 (e) 28583-237.13 69.91 Hzg) o Notices 0 130.58 222.96 HCI (g)92.3 95.27 186.69 0 Using the data in the table above, calculate the change in Gibbs free energy for the following reaction: H2(g) + Cl2(g) 2 HCl(g O-190.5 kJ/mol O-92.3kJ/mol O -95.27kJ/mol O -373.38 kJ/mol O-184.6 kJ/mol

Explanation / Answer

First we need to find delta Hrxn and delta Grxn and delta Srxn

Delta Hrxn = Hf of HCl - (Hf of H2 +Hf of Cl2)

= 2x(-92.3)- (0+0)= -184.6 kj/mole

Delta Srxn =2 x SO of HCl - (So of H2 +So of Cl2)

=2(186.69)-(130.59+222.96)= 373.39 - 353.55 = 19.84 kj/mol

Delta Grxn = Delta Hrxn - T Delta Srxn

= -184.6 x 103 - 298.15(19.84) = - 190.5 kj/mol

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