A 90.0 mL sample of 1.00M NaOH is mixed with 45.0mL of 1.00M H2SO4 in a large st

Question

A 90.0 mL sample of 1.00M NaOH is mixed with 45.0mL of 1.00M H2SO4 in a large styrofoam coffee cup; the cup is fitted with a lid through which passes a calibrated thermometer. The temperature of each solution before mixing is 21.1 Celsius. After adding the NaOH solution to the coffee cup and stirring the mixed solutions with the thermometer, the maximum tempreature measured is 31.5 Celsius. Assume that the density of the mixed solutions is 1.00g/mL, that the specific heat of the mixed solutions is 4.18J /(g * Celsius), and that no heat is lost to the surroundings.

Calculate the enthalpy change per mole of H2SO4 in the reaction.

I am not sure how to do this. Can you show me how to do it. Let me know if you want the points raised.

Explanation / Answer

Change in temperature is 30.5 - 23.1 degrees (either K or C) = 7.4 degrees
specific heat = 4.18 J/(g*C)
Over change in temperature this gives 4.18 * 7.4 J/g = 30.932 J/g
Using density get 30.932 J/ml

now all you need to do is work out volume in ml to obtain your answer.
2 NaOH + H2SO4 gives 2 NaSO4 + 2 H2O (? don't usually get 2 metal atoms in same molecule)
90ml of 1molar NaOH contains 0.09 moles of NaOH
45ml of 1molar H2SO4 contains 0.045 moles of NaOH
So total volume = 90ml + 45ml + volume of water created by reaction.
As I am not sure about my equation I will leave you to calculate the total volume
Enthalpy change(J) = total volume (ml) times 30.932 J/ml
I hope this helps.

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