In today\'s workshop we will use calorimetry to calculate the standard molar ent

Question

In today's workshop we will use calorimetry to calculate the standard molar enthalpy of reaction or the enthalpy change that occurs in a system when 1 mole of matter is transformed by a chemical reaction at 1 atm: AHEqron (in J/mol, at constant pressure and pV work only), where q cmAT Let's consider the following chemical experiment: A student wants to measure the heatof neutralization reaction. To do this she mixed in a coffer cup calorimeter 35.0 mL of 0.70MHCl solution and 65.0 mL of 0.50M NaoH solution both at 22.00C initially. She then continued to measure the temperature of the mixture for 10 minutes. 1. Write a chemical equation, a complete ionic and a net ionic equations for this reaction. Chemical equation Complete ionic equation Net ionic equation 2. Now discuss with your group the following questions: (a) What is(are) the product(s) of this reaction? (b) What is the system in this experimentand what are the surroundings? (c) Is this system an open one, a closed one or an isolated one? Before you write down your answers make sure that EVERYONE in your group can justify each them. Now write your answers below and don't forget to explain them. (a) is(are) the product(s) in this reaction. (b) The system: The surroundings: (c) We can consider this system to be an system, because

Explanation / Answer

(1)

Chemical equation: NaOH + HCl ----> NaCl + H2O

Complete ionic : Na+ + OH- + H+ + Cl- ----> Na+ + Cl- + H2O

Net ionic: H+ + OH- ----> H2O

(2)

(a)

Sodium chloride salt and water are the reaction products.

(b)

Solution is system

Calorimeter is surroundings

(c)

Isolated, because we are assuming no heat loss takes place

(3)

Moles of acid = 0.035*0.7 = 0.0245

Moles of base = 0.065*0.50 = 0.0325

Here, 1 mole acid reacts with 1 mole base to produce 1 mole water and salt each. So acid is in limiting amount

Thus, 0.0245 moles of water are produced

(4)

(a)

Over the 10 minutes duration, the temperature changes from 25.375 0C to 24.375 0C

Thus the temperature decrease equals = 25.375 - 24.375 = 1 0C

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.