Dissolving 4.02 g of cacl2 in enough water to make 349 ml of solution causes the temperature of the solution to increase by 3.91 oc. assume the specific heat of the solution and density of the solution are the same as water′s (about 4.18 j/goc and 1.00 g/cm3, respectively) calculate δh per mole of cacl2 (in kj) for the reaction under the above conditions.

ΔH per mole of CaCl2 is 160.3 kJ

Explanation:

Step 1: The balanced equation

CaCl2 (aq)→  Ca2+  +  2Cl-(aq)

Step 2: Data given

Molar mass of CaCl2 =110.98 g/mole

mass of CaCl2 = 4.02 grams

volume of the solution = 349 mL = 0.349 L

Temperature increases with 3.91 °C

Specific heat of the solution = specific hea of water = 4.18 J/°C*g

Density solution = density water = 1g/cm³

Step 3: Calculate moles of CaCl2

Numer of moles of CaCl2 = mass of CaCl2 / Molar mass of CaCl2

Number of moles = 4.02 grams /110.98 g/mole= 0.036 moles

Step 4 = Calculate amount of heat added

Q = m*c*ΔT

with m = the mass 349 grams + 4.02 grams = 353.02 grams

c = 4.18 J/°C*g

ΔT = 3.91 °C

Q = 353.02*4.18*3.91 = 5769.69 J

Step 5: Calculate ΔH per mole

5769.69 J /0.036 moles = 160269.1 J = 160.3 kJ

ΔH per mole of CaCl2 is 160.3 kJ

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