An experiment is run in 1000.00 mL of aqueous solution, where 0.500 mol are reac

Question

An experiment is run in 1000.00 mL of aqueous solution, where 0.500 mol are reached with an excess of acid. If the initial temperature of the reaction mixture was 24.00 degree C and the reaction had a Delta H of -458.5 kJ/mol, what is the final temperature of the reaction solution? Assume that there are no competing reactions, and that the solution's density and specific heat are identical to those of water (i.e., 1.00g/mL and 4.184J/g degree C. respectively). Was the reaction above exothermic or endothermic? Now suppose that instead of just one reaction, there was also a competing reaction going on in the same vessel the same time. If the competing reaction had a Delta H of 79 kJ/mol and consumed exactly 10% of the starting material, what would the final temperature the solution be? Assume that there were no further competing reaction, and that the products do not change the properties of the solution (i.e., same density and specify heat as water?

Explanation / Answer

Problem 1

Let's gather the data we have:

Vsol = 1000.00mL

nMetal = 0.500mol

T1 = 24.000°C

H = -854.5 kJ/mol

d = 1.00g/mL

C = 4.184 J/g°C

We'll use the specific heat formula:

q = -mCT

The mass m of the solution is calculated using its density and its volume:

d = m/v

m = dv

m = (1.00g/mL)(1000mL)

m = 1000g

While q is calculated using H and the number of moles of metal

q/n = H

q = nH

q = 0.5mol (-854.5 kJ/mol)

q = -427.25 kJ

We can now substitute q, m, Ce and T1 in the formula:

q = -mCT

-427,250J = -(1000g)(4.184 J/g°C)(T2 - 24°C)

102.115°C = T2 - 24°C

T2 = 126.115°C

So, the final temperature of the solution (considering no competing reactions) would be 126.115°C and the reaction would be exothermic because H is negative.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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