Math and Chemistry Review: You may work in groups of two students, but you will
ID: 473896 • Letter: M
Question
Math and Chemistry Review: You may work in groups of two students, but you will each turn in your own assignment. Chemistry Math: When suitable, provide on explanation for your answer, (revised for 4^th Ed. Tro) Drawing energy diagrams (figure 6.10 page 273. purple arrow): An energy diagram that shows the relative energies of the reactants and products. The energy difference between the reactants and products should be the reaction enthalpy for the balanced reaction, Delta H On the back of this page, calculate the reaction enthalpies for the following reactions and draw the energy diagrams: H_2O (l) - H_2(g) + O_2 (g) (unbalanced) CO_2(g) H_2O (l) rightarrow C_6H_120_6 (s) + O_2 (g) (unbalanced) Examine figure 14.12 & 14.13 on page 642-643 of your text. Compare this diagram to the diagram you have drawn on the back of this page. What do the differences represent? Stoichiometry problems. For the reaction given in problem 1 b) a. How many grams of glucose (C_6 H_12 0_6) are produced with the production of 125 mL of oxygen gas (255 mmHg, 25degree C)? How many mL of water are used in this process? Assume a density of 0.998 g/mL. Complete the net ionic equations for the following pairs of reactants (pg. 167, Example 4.12). The solubility rules can be found on page 161 (or use the last semesters handout) BaS(aq) + CuS0_4(aq) rightarrow Cr(OH)_3 (s) + HBr MgBr_z (oq) + ZnSO_4 (oq) rightarrowExplanation / Answer
for the reaction, 2H2O (l)---->2H2(g)+O2(g)
enthalpy change = 2* deltaH for H2+1* enthalpy of O2- 2* enthalpy of H2O
since delta=0+0-(2*(-285.8)= 571.6 Kj/mole. Endothermic reaction.
for the 2nd reaction, 6CO2(g)+ 6H2O(l) -----> C6H12O6+ 6O2(g)
enthalpy change = enthalpy of C6H12O6+ 6* enthalpy of O2- ( 6* enthalpy of CO2 + 6* enthalpy of H2O(l)
=-1273.02 +0 - {(6*(-393.5) + 6*(-285.8) }=2802.78 Kj/mole. This is also endothermic reaction. Hence energy diagram is
2. Moles of oxygen produced can be calcualted from gas law equation,
V= 125ml =125/1000 =0.125L, P= 255mm Hg= 255/760 atm =0.3355atm, T= 25 deg.c= 25+273=298K,
R= 0.0821 L.atm/mole.K, from PV=nRT
n= PV/RT= 0.3355*0.125/ (0.0821* 298)= 0.001714 moles
from the reaction 6CO2+ 6H2O---------> C6H12O6+ 6O2
6 moles of oxygen along with 1 mole of glucose and requires 6 moles of water
moles of glucose = 0.001714/6 =0.000286 moles. Mass of glucose= moles* molar mass
molar mass of glucose (C6H12O6)= 6*12+1*2+6*16= 180, mass of glucose= 0.000286*180 =0.0514 gm
moles of water= 0.001714. mass of water = moles* molar mass =0.001714*18= 0.030854 gm
volume of water= mass/ density =0.030854/0.998 ml=0.030916ml
3. Bas(aq)+CuSO4(aq) ----> BaSO4 +CuS(s)
Ba+2 +S2- + Cu+2 + SO4-2 ------->Ba+2 + SO4-2 +Cus(s) ( ionic equation)
Cu+2+S2- ------>CuS ( net ionic)
b) Cr(OH)3+ 3HBr---->CrBr3 (aq)+ 3H2O(l)
Cr(OH)3 + 3H+ + 3Br- ---------->Cr+3+3Br- + 3H2O(l)
Cr(OH)3+ 3H+ -------> Cr+3 + 3H2O(l)
c. MgBr2+ZnSO4---> no net ionic equation