CHEM11041 Chemistry for the Life Sciences

1. Write a balanced equation for the reaction. 
2. Is this reaction endothermic or exothermic Explain. 
3. Write an expression for the equilibrium constant for this reaction. 
4. Given that the value of the equilibrium constant is 48; would you expect this reaction to be fast or slow Explain
5. Explain the effect on equilibrium of the formation of HI form H₂reaction with I₂ by;
6. Increasing temperature 
7. Increasing pressure by decreasing the volume 
8. Decreasing concentration of hydrogen 
9. Increasing the concentration of iodine 
10. Adding a catalyst 

 Sugar cane may be used as a renewable source of ethanol, C₂H₅OH, via the conversion of sucrose, C₁₂H₂₂O₁₁, according to the following unbalanced equation:

1. Explain why this is a chemical reaction. 
2. Balance the equation for this reaction. 
3. How many grams of ethanol are formed from the complete conversion of 0.750 moles of sucrose
4. How many grams of ethanol are formed from the complete conversion of 37.5g of sucrose
5. What is the theoretical yield of ethanol in grams for the complete conversion of 57.5 g of sucrose
6. If 25.0 g of ethanol actually forms, what is the percentage yield of ethanol for the reaction of 57.5 g of sucroseDescribe how you would prepare mL of a 0.250 M solution of nitric acid, HNO₃, available commercially at a concentration of 16.0 M.

500. Describe how you would prepare 250.0 mL of a 0.050 M sucrose solution given that the sucrose is supplied as a solid. Molar mass of sucrose is 342.2965 gmol^-1  
501. Which lowers the freezing point of 2.0 kg of water more, 0.20 moles of NaOH or 0.20 moles of Ba(OH)₂ Explain your answer. 

Hydrogen iodide gas, HI, forms when hydrogen, H₂ and iodine, I₂ react.

   = + 26.5 kJ/mole

• Balance equation for the reaction 
• The reaction would be “endothermic” because  of the reaction is positive (
• Expression for the equilibrium constant for this reaction
• Given value of equilibrium constant is , it is apparent that the value is high and therefore, the product concentration would be higher than the reaction concentration. Hence, the reaction would take place very fast.
• Effects on equilibrium of the formation of HI from  reaction with by the following cases:
• Increasing temperature– According to the Le chatelier principle, when the temperature of endothermic reaction increases the reaction equilibrium would favour to shift in the right side direction (forward direction). Therefore, increase in the reaction temperature would increase the formation of HI.
• Increasing pressure by decreasing the volume - According to the Le chatelier principle, increase in the reaction pressure would shift the reaction equilibrium in that direction on which number of moles are low. In the present case, number of moles of reactants and products in both the direction is same. Therefore, increase in pressure would not affect the equilibrium of reaction.
• Decreasing concentration of hydrogen- From the equilibrium reaction equation, it is apparent that decease in the concentration of hydrogen would shift the reaction equilibrium in the left side (backward) direction.
• Increasing the concentration of iodine- From the equilibrium reaction equation, it is apparent that decrease in the concentration of iodine would shift the reaction equilibrium in the right side (forward) direction.
• Adding a catalyst– Addition of catalyst in the reaction would not affect the reaction equilibrium.

Unbalanced equation for the formation of ethanol

• Chemical reaction

When two or more reactant with different chemical composition forms a product of different chemical composition then it would be termed as chemical reaction. From the above reaction, it is apparent that sucrose and water are mixed together to form ethanol and hence, it is a chemical reaction.

• Balance the equation for the reaction
• Complete conversion  = 0.750 moles of sucrose

Grams of ethanol formed

Molar mass ethanol

From the reaction,

1 mole of sucrose produces = 4 moles of ethanol

1 mole of sucrose produces 4 moles of ethanol = 4 * 46 grams of ethanol

0.750 moles of sucrose would forms grams of ethanol = (0.750 * 4 * 46)/1 = 138 grams

• Complete conversion  = 37.5 grams of sucrose

Grams of ethanol formed

Molar mass sucrose

1 mole (342 grams) of sucrose would produce 4 moles of ethanol = 4*46 grams of ethanol

37.5 grams of sucrose would produce grams of ethanol = (4*46*37.5)/342 = 20.2 grams

• Complete conversion  = 57.5 grams of sucrose

Theoretical yield of ethanol in grams

1 mole (342 grams) of sucrose would produce 4 moles of ethanol = 4*46 grams of ethanol

57.5 grams of sucrose would produce grams of ethanol = (4*46*57.5)/342 = 30.93 grams

Therefore, theoretical yield of ethanol in grams is 30.93 grams.

• Grams of ethanol formed (Actual yield) = 25.0 grams

Percentage of yield of ethanol for the reaction of 57.5 gram of sucrose

Percentage of yield of ethanol for the reaction

• Helium balloon has a volume   

Volume ( )  

Idea Gas law   

Here, 

Milligrams of 0.80% w/v nalorphine solution

Obtained dose = 2.5 mg

0.80 % w/v means,

 0.80 grams of solute in 100 ml

2.5mg of solute = (100*2.5*10^-3)/0.80 = 0.3125 ml

• London Dispersion forces  - Non- polar molecular possess

Dipole – dipole forces – Molecules with net dipole moment possess

Hydrogen bonding – Molecule contains -OH groups possess

From the structure of dimethyl ether and ethanol, the intermolecular forces can be seen as

Intermolecular forces in dimethyl ether (CH₃OCH₃)   = Dipole- dipole forces

Intermolecular forces in ethanol (C₂H₅OH) = hydrogen bonding

Hydrogen bonding intermolecular force is higher than dipole- dipole force and hence, the boiling point of ethanol would be greater than boiling point of dimethyl ether.

• Preparation volume = 500 ml

Molarity of solution of nitric acid = 0.250 M

Concentration (commercially available)

Volume of nitric acid

Hence, in order to prepare the solution of 500 ml volume of the commercially available concentration of nitric acid, 7.8125 ml of 16M of nitric acid needs to be diluted up to a volume of 750ml.

• Preparation volume

Molarity of solution = 0.050 M

Molar mass of sucrose = 342.2965 per gram mole

Therefore, the mass of sucrose required 4.27 grams to prepare the solution.

Therefore, in order to prepare the solution of 250ml volume, 4.28 grams of sucrose needs to add in 250 ml of water to get the desired molarity of solution i.e. 0.05M.

• Need to determine, which of the compound would lower the freezing point of 2 kg of water.
• 20 moles of NaOH
• 20 moles of Ba(OH)₂

Lowering of freezing point of a solvent is directly proportional to the number of solute molecules (moles or ions). Hence, the compound which has maximum number of solute ions would lower the freezing point of 2kg of water.

Dissolution of compounds

  (Total ions after dissolution 2)

   (Total ions after dissolution 3)

It is apparent that  would provide 3 ions after dissolution and therefore, in order to lower the freezing point of 2 kg of water, the best suitable method is to use 0.20 moles of Ba(OH)₂