Q1) Suppose that 23 g of each of the following substances is initially at 27.0 C

Question

Q1)

Suppose that 23 g of each of the following substances is initially at 27.0 C. What is the final temperature of each substance upon absorbing 2.45 kJ of heat?

Part A

gold

Part B

silver

Part C

aluminum

Part D

water

Q2) A 32.5 g iron rod, initially at 22.7 C, is submerged into an unknown mass of water at 63.1 C, in an insulated container. The final temperature of the mixture upon reaching thermal equilibrium is58.4 C.

What is the mass of the water?

Express your answer to two significant figures and include the appropriate units.

Q3)Instant cold packs, often used to ice athletic injuries on the field, contain ammonium nitrate and water separated by a thin plastic divider. When the divider is broken, the ammonium nitrate dissolves according to the following endothermic reaction:
NH4NO3(s)NH+4(aq)+NO3(aq)
In order to measure the enthalpy change for this reaction, 1.25 g of NH4NO3 is dissolved in enough water to make 25.0 mL of solution. The initial temperature is 25.8 C and the final temperature (after the solid dissolves) is 21.9 C.

Part A

Calculate the change in enthalpy for the reaction in kilojoules per mole. (Use 1.0g/mL as the density of the solution and 4.18J/gC as the specific heat capacity.)

Express your answer to two significant figures and include the appropriate units.

Q4)Zinc metal reacts with hydrochloric acid according to the following balanced equation.

Zn(s)+2HCl(aq)ZnCl2(aq)+H2(g)

When 0.107 g of Zn(s) is combined with enough HCl to make 54.4 mL of solution in a coffee-cup calorimeter, all of the zinc reacts, raising the temperature of the solution from 22.3 C to 24.3 C.

Part A

Find Hrxn for this reaction as written. (Use 1.0 g/mL for the density of the solution and 4.18 J/gC as the specific heat capacity.)

Explanation / Answer

Q1) given data

Mass of substance = 23 g

initial temperature T1 = 27.0 c

Heat absorbed = 2.45 kJ * 1000 J/ 1 kJ = 2450 J

Final temperature T2 = ?

Part A Gold

specific heat of gold 0.13 J/ gC

using the following formula we can calculate the final temperature of the gold

q=m*c*(T2-T1)

where q= heat , m= mass , c= specific heat , T1 and T2 initial and final temperature

lets put the values in the formula

2450 J = 23 g * 0.13 J per g C * (T2-27 C)

2450 J = 2.99 T2 – 80.73

(2450 +80.73) / 2.99 = T2

846 C = T2

Final temperature of gold = 846 C

Part B Silver

Specific heat of silver = 0.23 J per gC

Now lets put the values in the formula

2450 J = 23 g * 0.23 J per g C * (T2-27 C)

2450 J = 5.29 T2 – 142.83

(2450 +142.83) / 5.29 = T2

490 C = T2

Final temperature of Silver = 490 C

Part C Aluminum

Specific heat of aluminum = 0.91 J per g C

Lets put the values in the formula.

2450 J = 23 g * 0.91 J per g C * (T2-27 C)

2450 J = 20.93 T2 – 565.11

(2450 +565.11) / 520.93 = T2

144 C = T2

Final temperature of Aluminum = 144 C

Part D) water

Specific heat of water = 4.184 J per g C

Lets put the values in the formula

2450 J = 23 g * 4.184 J per g C * (T2-27 C)

2450 J = 96.232 T2 – 2598.264

(2450 +2598.264) / 96.232 = T2

52.5 C = T2

Final temperature of water = 52.5 C

Q2 Given data

Mass of iron rod = 32.5 g

Initial temperature = 22.7 c

Initial temperature of water = 63.1 C

Final temperature = 58.4 C

Mass of water = ?

Change in temperature of water = 58.4 C-63.1 C = -4.7 C

Change in temperature of iron rod = 58.4 C – 22.7 C =35.7 C

Specific heat of water = 4.184 J per gC

Specific heat of iron = 0.45 J per g C

Now here water is losing heat and iron is absorbing heat

So we can make the following set up

-q water = q iron rod

- m*c*delta T = m*c*delta T

Lets put the values in the formula

-m*4.184 J per gC * (-4.7C) = 32.5 g * 0.45 J per g C * 35.7 C

m*19.6648 = 522.1125

m=522.1125/19.6648

m=26.6 g water.

Therefore mass of water = 26.6 g

Q3 Given data

Mass of NH4NO3 = 1.25 g

Volume = 25.0 ml = 25.0 g (because density is 1 g / ml)

Initial temperature = 25.8 C

Final temperature = 21.9 C

Change in temperature = 25.8 C -21.9 C = 3.9 C

Specific heat of solution = 4.184 J per g C

Now lets calculate the amount of heat absorbed by the reaction

q=m*c*delta t

lets put the values in the formula

q= 25.0 g * 4.184 J per gC * 3.9 C

q = 407.6 J

so now lets convert it to kJ per mol

(407.6 J * 1 kJ / 100 J) * (80.0434 g/ 1.25 g) =26.1 kJ/ mol

So the enthalpy change for the NH4NO3 reaction is 26.1 kJ/ mol

Q4 Given data

Mass of Zn = 0.107 g

Volume = 54.4 ml = 54.4 g (density = 1 g/ml)

Specific heat of water = 4.18 J per g C

Initial temperature = 22.3 C

Final temperature = 24.3 C

Change in temperature = 24.3 C -22.3 C = 2.0 C

Specific heat of solution = 4.184 J per g C

Now lets calculate the amount of heat absorbed by the reaction

q=m*c*delta t

lets put the values in the formula

q = 54.4 g * 4.18 J per g C * 2.0 c

q= 454.8 J

now lets convert this energy from Joules to KJ and and per mole of Zn

(454.8 J* 65.39 g /0.107 g)*(1 kJ / 1000 J) = 278 kJ/mol

Therefore enthalpy change for the reaction = 278 kJ / mol

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