When 5.11 g NaOH(s) is reacted with 200.00 ml of 0.639 M HCl(aq), the temperature increases 15.28 Celsius degrees. NaOH(s) + H+(aq) -> Na+(aq) + H2O(l) Find delta H.
`NaOH (s) + H^+ -> Na^+ (aq) + H_2O (l)`
First, we have to determine the amount of the reactants in terms of moles.
`5.11 grams NaOH * (1 mol e NaOH)/(40 grams)`
= 0.12775 moles `NaOH`
`0.639 (mol es HCl)/(L) * (200/1000) L`
= 0.1278 moles HCl = moles `H^+`
Essentially, there are equal amounts of reactants.
To know the amount of heat produced we should first treat the reaction to be in a calorimeter so as to assume that there is heat loss. The water molecules that will be produced will absorb the heat in the system thus the change in temperature will appear. We know that the heat capacity of water is 4.184 Joules/g degree C, by multiplying it with the mass of water and the change in temperature, we will able to get the value of the heat (or `Delta H` ).
Mass of Water:
`0.1278 mol es H^+ * (1 mol e H_2O)/(1 mol e H^+) * (18 grams H_2O)/(1 mol e H_2O)`
= 2.3004 grams `H_2O`
Heat of the reaction:
`Delta H = 2.3004 grams H_2O * 4.184 (Jou l es)/(gram* ^o C) * (15.28 ^o C)`
`Delta H` = 147.1 Joules