A 100.0-mL sample of 1.00 M NaOH is mixed with 50.0 mL of 1.00 M H_2 SO_4 in a l

Question

A 100.0-mL sample of 1.00 M NaOH is mixed with 50.0 mL of 1.00 M H_2 SO_4 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 degree C. After adding the NaOH solution to the coffee cup and stirring the mixed solutions with the thermometer, the maximum temperature measured is 32.5 degree C. 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 degree C), and that no heat is lost to the surroundings. (a) Write a balanced chemical equation for the reaction that takes place in the Styrofoam cup. 2NaOh + H_2 SO_4 rightarrow Na_2 SO_4 + 2H_2 O (b) Is any NaOH or H_2 SO_4 left in the Styrofoam cup when the reaction is over? Yes No (c) Calculate the enthalpy change per mole of H_2 SO_4 in the reaction. Include phases in the balanced chemical equation. Type an open parenthesis "(" to add a phase. Phases should not be subscripted. Use the left and rightarrow keys to move the cursor out of a superscript or subscript in the module.

Explanation / Answer

a)

The chemical Equation

ratio is 1:2 so

2NaOH + H2SO4 = 2H2O + Na2SO4

b)

mmol of acid = MV = 50*1 = 50 mmol of H2SO4

mmol of base = MV = 100*1 = 100 mmol of NaOH

ratio is 1:2 so, 50 mmol of H2SO4 react with 100 mmol of NaOH

there is NO speices left, this si stoichiometrically correct

c)

Find dH per mol of H2SO4

Q = - HRxn

HRxn = -Q = -m*C*(Tf-Ti)

mass = 100+50 = 150 mL = 150 g

Hrxn - 150 * 4.18 * (32.5-21.1) = -7147.8 J

HRxn per mol of H"SO4 = HRxn / mol = - 7147.8 / 50 = -142.9 kJ/mol

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