when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.
the equation for the reaction is:
naoh + hcl → nacl + h₂o
moles of hcl = 0.0250 l hcl ×
0.700
mol hcl
1 l hcl = 0.0175 mol hcl
volume of solution = (25.0 + 25.0) ml = 50.0 ml
mass of solution = 50.0 ml soln ×
1.00
ml soln
= 50.0 g soln
δt=t2-t1
= (22.1 – 20.0) °c = 2.1 °c
the heats involved are
heat from neutralization + heat to warm solution + heat to warm calorimeter = 0
q1+q2+q3=0
nδh+mcδt+cδt=0
0.0175 mol ×
δh
+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0
0.0175 mol ×
δh
+ 439.32 j + 585.9 j = 0
0.0175 mol ×
δh
= -1025.22 j
δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol
