A 51.0 mL sample of 1.0 M NaOH is mixed with 52.0 mL of 1.0 M H_2SO_4 in a large

Question

A 51.0 mL sample of 1.0 M NaOH is mixed with 52.0 mL of 1.0 M H_2SO_4 in a large styrofoam coffee cup: the cup is fitted with a through which passes a calibrate therometer. The temperature of each solution before mixing is 19.7 degree C. After adding the NaOH solution to the coffee cup, the mixed solutions are starred until reaction is complete. Assume that the density of the mixed solutions is 1.00 g/mL, that the specific heat of the mixed solutions is 4 18 j/(g middot c), and that no heat is lost to the surroundings. The Delta H for the neutralization of NaOH with H2SO4 is - 114 kj/mol H2SO4. What is the maximum measured temperature in the Styrofoam Cup?

Explanation / Answer

2NaOH(aq) + H2SO4(aq) --> Na2SO4(aq) + 2H2O(l)

Moles NaOH: 0.051 L x 1.00M = 0.051 moles base
Moles H2SO4: 0.052 L x 1.00M = 0.052 moles acid

If no heat lost, q = mass x specific heat x delta T
mass = volume*density = (51 mL + 52 mL)*1.00 g/mL = 103mL solution x 1.00g/mL = 103g
q = 103g x 4.18 J/g oC x (T- 19.7)

Given, q = 114000 J/mol of H2SO4 = 114000*0.052 J = 5928 J

=> T = (5928/(103*4.18)) +19.7 = 33.47oC

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