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Get Full Access to Chemistry: A Molecular Approach - 3 Edition - Chapter 9 - Problem 110e
Get Full Access to Chemistry: A Molecular Approach - 3 Edition - Chapter 9 - Problem 110e

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# Calculate of HBr(g) is not equal to one-half of the value

ISBN: 9780321809247 1

## Solution for problem 110E Chapter 9

Chemistry: A Molecular Approach | 3rd Edition

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Problem 110E

Calculate o for the reaction H2(g) + Br2(g)  2 HBr(g) using the bond energy values. The  of HBr(g) is not equal to one-half of the value calculated. Account for the difference.

Step-by-Step Solution:

Solution 110E Step 1 of 1 For the reaction We can state that one HH bond and one Br-Br bond break up, while two HBr bond form. We have to remember that when a reaction happens, we have to spend energy to break bonds (positive H ), while broken bonds release energy (negative H) Thus since For H-H H = 436 kJ/mol For Br-Br H = 193 kJ/mol Thus H broken bonds= (H H-H)+ (H Br-Br 629kJ/mol For H-Br H = - 364 kJ/mol Thus H formed bonds= 2(H H-Br= -728 kJ/mol Since H RXN (H broken bonds) + (H formed bonds) H = (629 kJ/mol) + (- 728 kJ/mol)= -99 kJ/mol RXN However, standard heat of formation (H ) is -36.3 kJ/mol at 25 °C. This descrepance between 0 the H (theoretical value) and H (measured value) is because the theoretical value doesn’t take in account of correction factors like the heat of condensation of fusion to achieve the standard state conditions.

Step 2 of 1

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