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Chemistry LibreTexts

8.12: Solutions for Selected Problems.

  • Page ID
    200195
  • Bonds Broken:

    \(C-C \: 6 \times 80 \frac{kcal}{mol} = 480 \frac{kcal}{mol}\)

    \(C-H \: 16 \times 100 \frac{kcal}{mol} = 1600 \frac{kcal}{mol}\)

    \(O=O \: 7 \times 120 \frac{kcal}{mol} = 840 \frac{kcal}{mol}\)

    Total: 2,920 kcal/mol

    Bonds Made:

    \(C=O \: 14 \times (-190 \frac{kcal}{mol}) = -2660 \frac{kcal}{mol}\)

    \(O-H \: 16 \times (-110 \frac{kcal}{mol}) = -1760 \frac{kcal}{mol}\)

    Total: -4,420 kcal/mol

    Overall: \(1240- 4420 \frac{kcal}{mol} = -1500 \frac{kcal}{mol}\)

    Bonds Broken:

    \(C-C \: 7 \times 80 \frac{kcal}{mol} =560 \frac{kcal}{mol}\)

    \(C-H \: 18 \times 100 \frac{kcal}{mol} = 1800 \frac{kcal}{mol}\)

    \(O=O \: 12.5 \times 120 \frac{kcal}{mol} = 1500 \frac{kcal}{mol}\)

    Total: 3,860 kcal/mol

    Bonds Made:

    \(C=O \: 16 \times (-190 \frac{kcal}{mol}) = -3040 \frac{kcal}{mol}\)

    \(O-H \: 18 \times (-110 \frac{kcal}{mol}) = -1980 \frac{kcal}{mol}\)

    Total: -5,020 kcal/mol

    Overall: \(3860 - 5020 \frac{kcal}{mol}= -1160 \frac{kcal}{mol}\)

    Bonds Broken:

    \(C-C \: 6 \times 80 \frac{kcal}{mol} = 480 \frac{kcal}{mol}\)

    \(C-H \: 7 \times 100 \frac{kcal}{mol} = 700 \frac{kcal}{mol}\)

    \(C-O \: 7 \times 85 \frac{kcal}{mol} = 595 \frac{kcal}{mol}\)

    \(O-H \: 5 \times 110 \frac{kcal}{mol} = 550 \frac{kcal}{mol}\)

    \(O=O \: 6 \times 120 \frac{kcal}{mol} = 840 \frac{kcal}{mol}\)

    Total: 3,165 kcal/mol

    Bonds Made:

    \(C=O \: 12 \times (-190 \frac{kcal}{mol}) = -2280 \frac{kcal}{mol}\)

    \(O-H \: 12 \times (-110 \frac{kcal}{mol} = -1320 \frac{kcal}{mol}\)

    Total: -3,600 kcal/mol

    Overall: \(3165 - 3600 \frac{kcal}{mol} = -435 \frac{kcal}{mol}\)

    GLATPtoADPmech.png

    In the mechanism for hydrolysis, water acts as a nucleophile and ATP acts as an electrophile. That's a problem because ATP is negatively charged. It will not attract electrons very easily. By binding to magnesium ion (Mg2+), the charge on the ATP will be lowered, accelerating the reaction with water.

    \(glucose + 2ATP \rightarrow 2G3P + 2ADP\)

    \(G3O + NAD^{+} + PO_{4}^{3-} + 2ADP \rightarrow pyr + NADH + 2ATP + H_{2}O\)

    First we need to realise that one glucose gives rise to two molecules of G3P, so the second phase occurs twice for every glucose molecule consumed.

    \(2G3P + 2NAD^{+} + 2PO_{4}^{3-} + 4ADP \rightarrow 2pyr + 2NADH + 4ATP + 2H_{2}O\)

    Adding the equations for the two phases together gives:

    \(glucose + 2ATP + 2G3P + 2NAD^{+} + 2PO_{4}^{3-} + 4ADP \rightarrow 2G3P + 2ADP + 2pyr + 2NADH + 4ATP + 2H_{2}O\)

    That equation can be simplified, because some things appear on both the left and the right. It's just like algebra.

    \(glucose + 2NAD^{+} + 2PO_{4}^{3-} + 2ADP \rightarrow 2pyr + 2NADH + 2ATP + 2H_{2}O\)

    GLGluccyclizeQ.png
    GLbetaglucose.png
    GLacetalsoln.png
    GLpyranosesoln.png
    GLfuransoln.png
    GLGtoG6Pmech.png
    GLG6Ptoenolmech.png
    GLenoltoF6Pmech.png
    GLenolsoln.png
    GLketosoln.png
    GLAldolpredictsoln.png
    GLretroaldolpredsoln.png
    GLFBPtomixmech.png
    GLDHAPtoG3Pmech.png
    GLaldolcatsoln.png
    GLaldolcatsoln.png
    GLretroaldolcatsoln.png