2. 50.0 mL of a solution of HCl is combined with 100.0 mL of 1.15 M NaOH ma calo
ID: 812451 • Letter: 2
Question
2. 50.0 mL of a solution of HCl is combined with 100.0 mL of 1.15 M NaOH ma calorimeter. The reaction mixture is initially at 22.4 degreeC and the final temperature after reaction is 3 1.2 degreeC. What is the molarity of the HCl solution? You may assume that there is an excess of base (so all of the HCl has reacted), that the specific heat of the reaction mixture is 0.96 cal/g degreeC, and that the density of the reaction mixture is 1.02 g/mL. The heat of neutralization of HCl and NaOH is 13.6 kcal/mol. 7. Given the following thermochemical equations: N2 (g) + O2(g) - > 2 NO (g) deltaH = + 180.75 kJ N2 (g) + 3 H2 (g) - > 2 NH3 (g) deltaH = -92.47 kJ 2 H2 (g) + O2(g) - > 2 H2O (l) deltaH = -483.67 kJ calculate delta H for the following thermochemical reaction: 4 NH3 (g) + 5O2(g) - > 4 NO (g) + 6 H2O (l)Explanation / Answer
HCL + NaOH = NaCl + H2O delta H = 13.6 kacl/mole of HCl reacted
Total volume of the reaction mixture = 50 + 100 = 150 ml
density of the reaction mixture = 1.02 g/ml
so mass = density * volume ==> 1.02 * 150 = 153 g
Q = mC dtT
153 g * 0.96 cal/g * C * (31.2 - 22.4)0 C
= 1292.544 cal
= 1.293 kcal
Since its given as 13.6 kacl/mole of HCl reacted,
1.293 kcal * 1 mole HCl = 0.0950 moles of Hcl has been consumed
1 13.6 kcal
molarity of HCl = 0.0950 moles * (0.150 litre = total volume) = 0.0143 M
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equation 2 reversing and multiplying with 2
4NH3 -----------------> 2N2 + 6H2 delta H = 2 * (+92.47) = +184.94 KJ
multiply eq 1 with 2
2N2 + 2O2 --------> 4 NO delta H = 2 (180.75) = 361.50 KJ
eq 3 * 3 times
6H2 + 3O2 ------------> 6H2O delta H = 3 (-483.67) = - 1451.01 KJ
Adding all these 3 new equations, we get
4NH3 -----------------> 2N2 + 6H2 delta H = 2 * (+92.47) = +184.94 KJ
2N2 + 2O2 --------> 4 NO delta H = 2 (180.75) = 361.50 KJ
6H2 + 3O2 ------------> 6H2O delta H = 3 (-483.67) = - 1451.01 KJ
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4NH3 + 5O2 ----------> 4NO + 6H2O delta H = - 904.57 KJ
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