Which of t he three AH\'s was the largest? Explain briefly! 1. 2. In reaction 1,
ID: 961895 • Letter: W
Question
Which of t he three AH's was the largest? Explain briefly! 1. 2. In reaction 1, if we had used 4 grams of NaOH instead of 2 grams, would the heat gained by the solution [# 6] have been larger, smaller or the same? Explain briefly! 3. In reaction 2, if we had used 4 grams of NaOH instead of 2 grams, would the heat gained by the solution [# 6] have been larger, smaller or the same? Explain briefly! We ignored the heat lost to the Styrofoam cup and the air in the calculations for this lab. Did this caus your calculated values for aH [# 8] to be too large or too small. Explain briefly! 4. 5. Calculate the heat of neutralization for limewater, Ca(OH)(aq), and hydrochloric acid, HCI Ca(OH),(aq) + 2 HCI(aq) CaCl,(aq) + 2 H,01) using Hess's Law and the three thermochemical equations below. Show work. r--186 kJ (I) CaO(s) + 2 HCI(aq) CaCL(aq) + H,0(1) (2) CaO(s) + H2O(l) Ca(OH),(s) (3) Ca(OH)(8)Ca(OH)(aq)Explanation / Answer
1. The enthalpy change for reaction 3 is larger and. this is due to significant change in temperature.
2. when heat gained remains the same since it is expressed in terms of Kj/mol
3.since heat gained is expresed in Kj/mol. it remains the same.
4. Since heat is lost, this will decrease the enthalpy change. since the heat capacity of styorofoam is very less, the decrease is not significant.
5.
Reversing Eq.3 gives it gives
Ca(OH)2(aq)---->Ca(OH)2 (s) Enthalpy change= 13 Kj (3A)
Reversing Eq.2 gives
Ca(OH)2(s) --> CaO(S)+ H2O(l) Enthalpy change= 62 Kj (2A)
Addition of Eq.3A and 2A gives
Ca(OH)2(aq)---> CaO(s)+ H2O(l) enthalpy change= 62+13= 75 KJ (4)
Eq.1 is CaO(s)+ 2HCl(aq)---> CaCl2(aq)+ H2O(l) enthalpy change= -186 KJ (1)
Addition of Eq.4 and Eq.1 gives
Ca(OH)2 (aq)+ 2HCl (aq)---->CaCl2(aq)+ 2H2O(l) enthalpy change= =186+75=-101Kj
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