# 8.12: Solutions for Selected Problems.

Bonds Broken:

$C-C \: 6 \times 80 \frac{kcal}{mol} = 480 \frac{kcal}{mol} \nonumber$

$C-H \: 16 \times 100 \frac{kcal}{mol} = 1600 \frac{kcal}{mol} \nonumber$

$O=O \: 7 \times 120 \frac{kcal}{mol} = 840 \frac{kcal}{mol} \nonumber$

Total: 2,920 kcal/mol

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

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

Total: -4,420 kcal/mol

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

Bonds Broken:

$C-C \: 7 \times 80 \frac{kcal}{mol} =560 \frac{kcal}{mol} \nonumber$

$C-H \: 18 \times 100 \frac{kcal}{mol} = 1800 \frac{kcal}{mol} \nonumber$

$O=O \: 12.5 \times 120 \frac{kcal}{mol} = 1500 \frac{kcal}{mol} \nonumber$

Total: 3,860 kcal/mol

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

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

Total: -5,020 kcal/mol

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

Bonds Broken:

$C-C \: 6 \times 80 \frac{kcal}{mol} = 480 \frac{kcal}{mol} \nonumber$

$C-H \: 7 \times 100 \frac{kcal}{mol} = 700 \frac{kcal}{mol} \nonumber$

$C-O \: 7 \times 85 \frac{kcal}{mol} = 595 \frac{kcal}{mol} \nonumber$

$O-H \: 5 \times 110 \frac{kcal}{mol} = 550 \frac{kcal}{mol} \nonumber$

$O=O \: 6 \times 120 \frac{kcal}{mol} = 840 \frac{kcal}{mol} \nonumber$

Total: 3,165 kcal/mol

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

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

Total: -3,600 kcal/mol

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

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.

$glucose + 2ATP \rightarrow 2G3P + 2ADP \nonumber$

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

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.

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

Adding the equations for the two phases together gives:

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

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

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