To propose a mechanism of mannose 6-phosphate derivative from fructose 6-phosphate. Concept introduction: Acid-base Catalysis by the enzyme. Step 1: Substrate binding. Step 2: Acid catalyzed ring opening. Step 3: Base abstracts the acidic proton from C2 to form a cis enediolete intermediate. Step 4: proten is replaced on C-1 as ab iverall proton teamster. Step 5: Ring disuse to form product.

Question

: At a certain temperature, the equilibrium constant K for the following reaction is 1.58 × 10-12 N2(g) + O2(g) = 2 NO(g) Use this information to complete the following table. Suppose a 38. L reaction vessel is filled with 0.93 mol of N2 and 0.93 mol of O2. What can you say about the composition of the mixture in the vessel at equilibrium? There will be very little N2 and O2. There will be very little NO. What is the equilibrium constant for the following reaction? Be sure your answer has the correct number of significant digits. 2 NO(g) N2(9)+02(9) What is the equilibrium constant for the following reaction? Be sure your answer has the correct number of significant digits. 3 N2(9)+302(g) 6 NO(g) Neither of the above is true. K = ☐ K = ☐ ☐ X10 Х D ? 000 18 Ar B

Answer 1

Question

Chapter 25.SE, Problem 31MP

Interpretation Introduction

Interpretation:

To propose a mechanism of mannose 6-phosphate derivative from fructose 6-phosphate.

Concept introduction:

Acid-base Catalysis by the enzyme.

Step 1: Substrate binding.

Step 2: Acid catalyzed ring opening.

Step 3: Base abstracts the acidic proton from C2 to form a cis enediolete intermediate.

Step 4: proten is replaced on C-1 as ab iverall proton teamster.

Step 5: Ring disuse to form product.

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