Question

The molar enthalpy change for the neutralization reaction: HCl (aq) NaOH (aq) NaCl (aq) H20 (I) is AH -57.3 kJ/mol. This is an exothermic reaction which means heat is released. The heat released by this reaction can be calculated with the equation q = nAH, where n is the number of moles of the limiting reagent. The heat released by this reaction increases the temperature of the reaction solution (the surroundings) Imagine you mix 50.00 mL of 1.00 M HCl with to 50.00 mL 1.00 M N2OH in a beaker at 23°C. Estimate the final temperature of the solution after the reaction has gone to completion. You will need to make a few assumptions: Assume no heat is lost to the environment. Assume the density of the reaction solution is the same as water, 1.00 g/mL Assume the specific heat capacity of the reaction solution is the same as of water, 4.184 J/g-deg.

Answer 1

Answer:

Final temperature is 29,8°C

Explanation:

For the reaction:

HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

You have 0,05000L*1,00M = 0,0500 moles of HCl ≡ moles of NaOH

The heat realeased by this reaction with these amounts is:

q = 57,3kJ/mol*0,0500mol = 2,865 kJ ≡ 2865J

The formula of heat capacity is:

q=C×ΔT×m

Where q is the heat released (2865 J)

C is specific heat capacity (4,184J/g°C)

m is mass (50 mL of NaOH+50 mL of HCl = 100 mL ≡ 100g -by density of solution=1,00g/mL-

ΔT is final T- initial T (X-23)

Replacing:

2865 J = 4,184J/g°C×100g×(X-23°C)

6,8 = X-23

X = 29,8°C

Final temperature is 29,8°C

I hope it helps!