Question

When a 12.8 g sample of KCL dissolves in 75.0 g of water in a calorimeter the temp. drops from 31 Celsius to 21.6 Celsius. Calculate deltaH for the process.

PLEASE EXPLAIN EVERY STEP.

Answer 1

To calculate the delta H for the dissolution of KCl in water, use the heat equation q = mcΔT with the given mass, specific heat capacity of water, and the temperature change. Then, convert q to kilojoules and divide by the number of moles of KCl to find the enthalpy change per mole.

Step-by-step explanation:

The question pertains to thermochemistry, a branch of chemistry that deals with the heat involved in chemical reactions and physical transformations. We need to calculate the enthalpy change (delta H) when KCl dissolves in water, which involves using the formula q = mcΔT, where q is the heat absorbed or released, m is the mass of the solvent (water), c is the specific heat capacity, and ΔT is the change in temperature.

First, we identify the temperature drop: ΔT = 21.6°C - 31.0°C = -9.4°C. This indicates that the solution absorbed heat (since the temperature of the water dropped, the reaction is endothermic). Then, using the specific heat capacity of water (4.18 J/g°C), we calculate q as follows:

q = mcΔT
= (75.0 g)(4.18 J/g°C)(-9.4°C)
= -2969.4 J

Since specific heat capacity and mass are both positive, the negative q value signifies that the heat was absorbed by the substance from the water. To find delta H in kJ/mol, we convert the mass of KCl to moles using its molar mass (74.55 g/mol) and then divide q by the number of moles:

Moles of KCl = 12.8 g / 74.55 g/mol
Approximately = 0.172 moles

Delta H = q / moles
= -2969.4 J / 0.172 mol
= -17261 J/mol

Since we need delta H in kilojoules per mole, we convert J to kJ:

Delta H = -17261 J/mol / 1000 J/kJ
= -17.261 kJ/mol

As the delta H is negative, it confirms that the dissolution of KCl in water is an endothermic process.

Answer 2

Step 1: Calculate qsur (the surrounding is

usually the water)

qsur = ? J

m = 75.0 g water

c = 4.184 J/g

oC

ΔT = (Tfinal- Tinitial)= (21.6 – 31.0) = -9.4 oC

qsur = m · c · (ΔT)

qsur = (75.0g) (4.184 J/g

oC) (-9.4 oC)

qsur = - 2949.72 J

First, using the information we know that we

must solve for qsur, which is the water. We know

the mass for water, 75.0g, the specific heat of

the water, 4.184 j/g

o

c, and the change in

temperature, 21.6-31.0 = -9.4 oC. Plugging it

into the equation, we solve for qsur.

Step 2: Calculate qsys qsys = - (qsur)

qsys = - (- 2949.72 J)

qsys = + 2949.72

In this case, the qsur is negative, which means

that the water lost energy. Where did it go? It

went to the system. Thus, the energy of the

system is negative, opposite, the energy of the

surrounding.

Step 3: Calculate moles of the substance

that is the system

Given: 12.8 g KCl

Mol system = (g system given)

(molar mass of system)

Mol system = (12.8 g KCl)

(39.10g + 35.45g)

Mol system = 12.8 g KCl

74.55 g

Mol system = 0.172

Here, we solve for the mol in the system by

using the molar mass of the material in the

system.

Step 4: Calculate ΔH ΔH = q sys .

Mol system

ΔH= + 2949.72 J

0.172 mol

ΔH= +17179.81 J/mol or +1.72 x 104

J/mol

i hope this helps