ASAP Orgo Chem 1 Chap 6 Klein 1. Given that ?G o for a reaction is -7.00 kcal mo
ID: 838410 • Letter: A
Question
ASAP Orgo Chem 1 Chap 6 Klein
1. Given that ?Go for a reaction is -7.00 kcal mol-1, what is the value of Keq at 25oC (i.e., 77oF)? The value of R is 0.008314 kJ/mol K or 0.001986 kcal/mol K. Report your answer to three significant digits.
2. Which of the following statements are true in regards to the following reaction coordinate diagram? (can be more than one answer)
3. Given that ?Go for a reaction is -6.00 kcal mol-1, what is the value of Keq at 25oC (i.e., 77oF)? The value of R is 0.008314 kJ/mol K or 0.001986 kcal/mol K. Report your answer to three significant digits.
4. An elephant has just had a sandwich that contained the equivalent of 9.00 mol of glucose (?Ho = -2810 kJ mol-1). The standard molar entropy (?So) of reaction is +182 J K-1 mol-1. How much energy (in kJ) is available for sustaining the elephant's muscular and nervous activity assuming a body temperature of 38.0oC? Make sure that your answer has the correct number of significant figures.
5.
6)
The overall reaction below is exergonic, and the transition from B to C is the rate-determining step. Which reaction coordinate diagram applies to this reaction?
7. Given that ?Go for a reaction is 7.00 kJ mol-1, what is the value of Keq at 25oC (i.e., 77oF)? The value of R is 0.008314 kJ/mol K or 0.001986 kcal/mol K. Report your answer to three significant digits.
8. Given that ?Go for a reaction is -1.00 kJ mol-1, what is the value of Keq at 25oC (i.e., 77oF)? The value of R is 0.008314 kJ/mol K or 0.001986 kcal/mol K. Report your answer to three significant digits.
9.
Which of the following statements are true in regards to the following reaction coordinate diagram?
d) The reaction will release more energy into the system than is consumed from the system; therefore, the reaction is exergonic.
10. Given that ?Go for a reaction is -6.00 kJ mol-1, what is the value of Keq at 25oC (i.e., 77oF)? The value of R is 0.008314 kJ/mol K or 0.001986 kcal/mol K. Report your answer to three significant digits.
a) The products are more stable than the reactants.Explanation / Answer
q1)
use the equation
?Go= -RTlogKeq
=> Keq=antilog(-?Go/RT)
where ?Go= 7 kcal/mol = 29288J/mol = 29.288kJ/mol
and T= 25 oC = 298 K
Keq = 3.547* 10^(-14)
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