Question

Consider 1.000 mol IBrs gas and the mixture of 1.500 mol Br2 +0.500 mol I2 at 1200 K. Assume that all vibrations contribute completely to the number of equipartition energy contributions R 8.314 JK-1mol1 What degrees of freedom are the most important in each example? Match the words in the left column to the appropriate blanks in the sentence on the right Reset Help vibrational The IBr3 has more degrees of freedom, whereas translational the Br2/I2 mixture has more degrees of freedom. Request Answer Submit Part B Using only the equipartition principle, estimate the difference in energy of the two substances, including vibrations, in J. Express your answer using three significant figures. VAcD EBrEBrt J

Answer 1

1. Moles of IBra 1.000 mel 2. Moles of Br2 – 1.500 mol & the 3. Moles of I. – 0.5 molt me. To 1200k ; R2 8. 314 J. K molto Since the higher vibrational degrees of freedom are activated only at higher temperature, they play a crucial role a to decide the energy of the gas molecules according to principle of equipartition of energy. At low temperature only ground skate or lower vibrational states are popl populated.

Now according to principle of equipartition of energes for a molecule containing Natoms, the following table is valid linear molecule. Nonlinear molecule No of degrees of freedom. Translational 3 3. Rotational 2 . 3 Vibrational. 3N-5 3.-6. So for I har vibration I Bra rotational & vibrational regrees of freedom : 3x4 -6 2 6 [since I Brz is non-linear] & for Br2 or I No of vibrational degrees of freedom: 3x2-5 – 1. Sie So Br₂ / Iz to mixture would have I +2 22 vibrational tegras of freedom & 3+3 .6 translational degrees of freedom.

boi IBT 1. I Br2 has more vibrational degrees of freedom By Br2 /I2 mixture has more whereas, translational degrees of freedom Past B According to equipartition principle of energy, Total energy for a linear molecule: {3+RT + 2x RT & (3N-5) *RT? J. mol & for non linear molecule : $ 3X RT + 3x RI + (3N:-6).*RT} I.no! where N is the no. of atoms in the molecule. So for IBro. f a d.ooo mol 3x 8131481200 to 8.21 + (3x4 - 6) x 8. 3 24x1200 } I mol 84807 8 I. 89 79L 2 I 8.31441200 2

co 2 For Brz Eer, o 1:500p mot § 3x 6314X1200 + 2 x 8-314X1200 & (3x2-5) x 8-3 24x1200 } Jimol a 1.500 mal x 34918.8 J. molle 52 378 2 J. I calculation for Timoll I2 20.500 x 34928.8J. for Brz & Iz would 2 17 459.4 J. T il I became Hence Erbre E Brzt Iz J - 89791.2 - 52 378.2 - 17459.4 19953.6 J utra