Nitric acid neutralizes potassium hydroxide. To determine the heat of reaction,

Question

Nitric acid neutralizes potassium hydroxide. To determine the heat of reaction, a student placed 63.8 mL of 3.5 M HNO3 in a coffee cup calorimeter, noted that the temperature was 27.9 °C, and added 63.8 mL of 3.5 M KOH, also at 27.9 °C. The mixture was stirred quickly with a thermometer, and its temperature rose to 39.5 °C.

Enter the balanced equation for the reaction. Do not include physical states.

Calculate the heat of reaction in joules. Assume that the specific heats of all solutions are 4.18 J·g-1·°C-1 and that all densities are 1.00 g·mL-1.

Calculate the heat of reaction per mole of acid (in units of kJ·mol-1).

Explanation / Answer

The heat of neutralization if found by using the equation q(neut) = mc(Tf - Ti)
c = heat capacity = 4.18 J/g°C; m = total mass of system = mass nitric acid solution + mass KOH;
number of mole of HNO3 = 3.5/1000*63.8 = 0.223 mole

mass of HNO3 = 0.223*63 g =14.06g

number of mole of KOH = 3.5/1000*63.8 = 0.223 mole

mass of KOH = 0.223*56 = 12.488g

total mass = 14.06g + 12.488g = 26.548g

q(neut) = (26.54 g)(4.18 J/g°C)(39.5 °C - 27.9 °C)

heat of reaction in joules=1286.87 J

The reaction is exothermic so the enthalpy will be negative.

H(neut) = - q(neut)/n where n = moles KOH

n = 0.223 mol

heat of reaction per mole of acid = H(neut) = - (1286.87 J)/(0.223 mol) = - 5770 J/mol = -5.77 kJ/mol

balanced equation  HNO3+ KOH -> KNO3 + H2O

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