Question

1. Consider this reaction:

H₂(g) +    O₂ (g)     à       H₂O (g)                                          dHo = __________

1. Balance the equation and put the correct number of moles in the equation above. (1 point)

2. Classify the forward reaction as (1) Combustion, (2) Decomposition, (3) Single replacement, (4) Double replacement, (5) Neutralization, (6) Synthesis. Circle the correct answer. (1 point)

3. In this reaction did Hydrogen get oxidized or reduced? Circle the correct answer. (1 point)

4. Based on your answer above is Hydrogen an oxidizing or reducing reactant? Circle the correct answer. (1 point)

5. In this reaction did Oxygen get oxidized or reduced? Circle the correct answer. (1 point)

6. Based on the answer above, is Oxygen an oxidizing or reducing agent? Circle the correct (1 point)
   
7. Calculate the dHo of the Synthesis reaction using the values from the table in Appendix 2. Show your work. (2 points)

8. If 4 moles of H2 is used in a balanced equation for the synthesis reaction, how many moles of water will be produced (without using the tables). Show your work. (1 point)

9. If 4 moles of H2 is used for the Synthesis reaction, calculate the enthalpy of the reaction without using the tables. Show your work. (1 point)

10. If 4 moles of H₂O is required to be produced, how many moles of H₂ and how many moles of O₂ will be required (without using tables). Show your work. (1 point).

11. If 1 mole of water is required to be produced, how many moles of H₂ and O₂ will be required (without using tables)? Show your work. (1 point)

12. If 6 moles of water is decomposed, how many moles of H₂ and O₂ will be formed? (1 point)

13. If 6 moles of water is decomposed, calculate the enthalpy of the reaction without using the values from the table available in Appendix 2 of your textbook. (2 points)

Answer 1

1)balanced rection:
2H2(g) + O2 (g) --> 2H2O (g)

2) type of reaction ; Synthesis
3)
Ox # of H in
H2 :: 0
H2O :: +1
Since ox # of H increased when product is formed, Hydrogen
is oxidized
4)
species undergone oxidation is reducing agent ;
Hence Hydrogen is reducing agent

5)
ox # of O in
O2 :: 0
H2O:: -2
Since ox # of O decreased from o to -2, oxygen is reduced
6)
Since oxygen under gone reduction,
oxidizing agent :oxygen

7)
2H2(g) + O2 (g) --> 2H2O (g)
values taken from standard table . [there may be slight
variation in different tables]

dHo =2* dH(H2O)f - 2* dH(H2)f - dH(O2)f
= 2* (-241.8) - 0 -0
dH0 = -483.6 kj
***********************
8)
2H2(g) + O2 (g) --> 2H2O (g)
2mol H2 produce 2 mole water
then 4 mole h2 produce 4 mol water
*********
9)
2H2(g) + O2 (g) --> 2H2O (g) . dH0 = -483.6 Kj
when 4 mole H2 used

dH0 = 4* -483.6 kj = -1934.4 Kj

***********
10)
for 1 mole water,
1 mole H2 and 1/2 mol O2 is needed

11)
From 1 mole water,1 mole H2 and 1/2 mol O2 is produced
From 6 mole water,6 mole H2 and 3 mol O2 is produced

12 )

from 2 mol water
dH0 = 483.6 mol

from 6 mol water

dH0 = 6/2 * 483.6 KJ = 1450.8 KJ

************
all solved . :))