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 56.5 mL of 3.9 M HNO3 in a coffee cup calorimeter, noted that the temperature was 25.3 °C, and added 56.5 mL of 3.9 M KOH, also at 25.3 °C. The mixture was stirred quickly with a thermometer, and its temperature rose to 32.3 °C. write the equation for the reaction and Calculate the heat of reaction per mole of acid (in units of kJ·mol-1. and 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

Explanation / Answer

Given :

Volume of HNO3 = 56.5 mL = 0.0565 L

[HNO3]= 3.9 M

Initial temperature = 25.3 deg C

Volume of KOH = 56.5 mL = 0.0565 L

[KOH]= 3.9 M

Final temperature = 32.3 deg C

Solution :

KOH (aq) + HNO3 (aq) --- > KNO3 (aq) + H2O (l)

Both are in equal volume and equal molarity.

Moles of KOH = moles of HNO3 = volume in L x molarity

= 0.0565 L x 3.9 M = 0.22035 mol

Heat of reaction Delta Hrxn = - q/ n

q = m c delta T

m is mass, C specific heat , Delta T is change in T.

Lets calculate mass of solution

Mass of solution = Total volume in mL x 1 g / mL

= (56.5 + 56.5 ) mL 1 g /mL

= 113 g

Lets plug given values in above equation.

q = 113 g x 4.184 J /(deg C g) x ( 32.3 – 25.3 ) deg C

= 3309.44 J

Delta H = - 3309.544 J/ 0.22035 mol

= 15019.50 J /mol

Delta H in kJ/mol

Delta H = 15.02 kJ/mol

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