When 50 ml of 1.5 M HCl is added to 100 ml of 1.5 M sodium hydroxide solution, temperature initially was 21.3 C and fainaly raised to 23.4 C, What is kj of heat produced? *

A)4643.2
mol
B)1102.9
mol
C)2211.05
mol
D)12381.86
mol

Q = -1.318 KJ

Explanation:

We will use the assumption that this solution acts like water and thus we will use the specific heat capacity of water and when converting from mL to g, we will use the conversion like we do for water.

We are told that 50 mL of HCl reacts with 100 mL of NaOH.

Thus total mass; m = 50 + 100 = 150 mL

Converting to grams gives 150 g since we have assumed that the solution behaves like water.

We are given;

Initial temperature; T_i = 21.3° C

Final temperature; T_f = 23.4° C

ΔT = 23.4 - 21.3

ΔT = 2.1°C

Formula for quantity of heat is;

Q = mcΔT

c is specific heat capacity.

We will use c = 4.184 J/g°C since the solution is assumed to behave like water.

Thus;

Q = -(150 × 2.1 × 4.184)

Q = -1317.96 J

Negative sign is used because temperature was raised and thus reaction is exothermic.

Approximation to KJ gives; -1.318 KJ

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