Question

Question #35c00

Answer 1

Here's what I got.

Explanation:

For starters, you know that the temperature of the solution increases as a result of the reaction

`21.00^@"C " -> " "24.70^@"C"`

so you can say that the enthalpy change of reaction will be negative--keep in mind that we use a minus sign to symbolize heat given off.

`DeltaH_ "rxn" = -"? kJ mol"^(-1)`

Now, the idea here is that the heat given off by the reaction will be absorbed by the solution.

To calculate the heat absorbed by the solution, you can use the following equation--we assume that the specific heat of the solution is equal to that of pure water

`color(blue)(ul(color(black)(q = m * c * DeltaT)))`

Here

• `q` is the heat gained by the solution
• `m` is the mass of the solution
• `c` is the specific heat of water, equal to `"4.18 J g"^(-1)""^@"C"^(-1)`
• `DeltaT` is the change in temperature, defined as the difference between the final temperature and the initial temperature of the sample

In your case, the change in temperature is equal to

`DeltaT = 24.70^@"C" - 21.00^@"C" = 3.70^@"C"`

This means that you have

`q_ "solution" = 144 color(red)(cancel(color(black)("g"))) * "4.18 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * 3.70 color(red)(cancel(color(black)(""^@"C")))`

`q_"solution" = "2227.1 J"`

You can thus say that when `2.00` moles of this unknown compound react with water, `"2227.1 J"` of heat are being given off to the surroundings.

The enthalpy change of reaction--expressed in kilojoules--will be equal to

`DeltaH_ "rxn for 2.00 moles" = -"2.23 kJ"`

When `2.00` moles of this unknown compound react, the reaction gives off `"2.23 kJ"` of heat.

We usually tend to express enthalpy changes in kilojoules per mole, you can say that the enthalpy change of reaction that occurs when `1.00` moles of compound react is equal to

`DeltaH_ "rxn for 1.00 mole" = (-"2.23 kJ")/"2.00 moles" = -"1.12 kJ mol"^(-1)`

The answers are rounded to three sig figs, the number of sig figs you have for the mass of the solution and for the number of moles of the compound.