A 100.0 mL sample of 0.300 M NaOH is mixed with a 100.0 mL sample of 0.300 M HNO

Question

A 100.0 mL sample of 0.300 M NaOH is mixed with a 100.0 mL sample of 0.300 M HNO3 in a coffee cup calorimeter. If both solutions were initially at 32.00 degree C and the temperature of the resulting solution was recorded as 37.00 degree C, determine the Delta H degree rxm (in units of kJ/mol NaOH) for the neutralization reaction between aqueous NaOH and HCI. Assume that no heat is lost to the calorimeter or the surroundings, and that the density and the heat capacity of the resulting solution are the same as water. -27.9 kJ/mol NaOH -139 kJ/mol NaOH -169 kJ/mol NaOH

Explanation / Answer

NaOH + HNO3 ---> NaNO3 + H2O

no of mol of NaOH = 100/1000*0.3 = 0.03 mol

mass of solution = 100+100 = 200 grams

heat released = 200*4.18*(37-32) = 4180 joule. = 4.18 kj

DHrxn = 4.18/0.03 = 139.33 kj/mol

       = -139.33 kj/mol

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