Using the standard enthalpies of formation listed in Table 1, compute the standa
ID: 1049747 • Letter: U
Question
Using the standard enthalpies of formation listed in Table 1, compute the standard enthalpy of reaction for the following reactions: Reaction 1: NaOH(s) rightarrow NaOH(aq) Reaction 2: NaOH(ag) + HCl (aq) rightarrow H_2O(l) + NaCl(aq) Table 1: Standard Enthalpies of Formation at 25 degree C from Atoms First. 2nd ed., McMurry and Fay Write the chemical equation for the reaction that results when Reaction 1 is combined with Reaction 2. This will be referred to as Reaction 3. Compute the standard enthalpy of reaction for the new reaction. How does this value compare to the standard enthalpies of reaction for Reaction 1 and Reaction 2?Explanation / Answer
answer 1 ) NAOH(s) ---> NAOH(liq) reaction 1
enthalpy of reaction = sum of the standard enthalpy of formation of product - sum of the standard enthalpy of fomation of product
Er = -470.1 -(-425.6) =+44.5 KJ/mole
NAOH (aq) + HCL (aq) --------> H2O (liq) + NACL (aq) reaction 2
enthalpy of reaction = sum of the standard enthalpy of formation of product - sum of the standard enthalpy of fomation of product
Er =-285.8 -240.1 - 167.2 -(-470.1 -167.2)
= -693.1 + 693.1 = -55.8 KJ/mole
answer2 ) on adding reaction we get
NAOH(s) + HCL(aq) ------> H2O (aq) + NACL(aq) reaction 3
answer 3) NAOH(s) + HCL(aq) ------> H2O (aq) + NACL(aq) for reaction 3
enthalpy of reaction = sum of the standard enthalpy of formation of product - sum of the standard enthalpy of fomation of product
Er = -240.1- 167.2 - 285.8 -(- 425.6 -167.2)
= -693.1 + 592.8 = -100.3 KJ/mole
reaction 1 =- 44.5 kj/mole , reaction2 = -55.8 kj/mole
reaction3 = reaction1 + reaction2
-100.3 kj/mole = -44.5 kj/mole - 55.8 kj/mole
-100.3 kj/mole = -100.3 kj/mole .
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