Question

Lithium bromide dissociates in water according to the following thermochemical equation: LiBr(s) → Li+ (aq) + Br– (aq) ΔH = –48.83 kJ/mol If 2.00 moles of lithium bromide are dissolved in 1000.0 grams of water at 25.0 °C, what is the final temperature of the water, assuming that all solutions have the same heat capacity as pure water (4.184 J/g-K)?

Answer 1

To find the final temperature of water after dissolving lithium bromide, we can use the equation q = mcΔT. The heat released by the dissolution of lithium bromide can be calculated using the given ΔH value. By substituting the values into the equation, we can find the change in temperature and determine the final temperature of the water.

Step-by-step explanation:

To find the final temperature of the water, we can use the equation q = mcΔT, where q is the heat absorbed or released, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.

First, calculate the heat released by the dissolution of 2.00 moles of lithium bromide. ΔH is given as -48.83 kJ/mol, so the heat released is -48.83 kJ/mol * 2.00 mol = -97.66 kJ.

Next, calculate the mass of water. 1000.0 grams of water is equivalent to 1000.0 mL, and since the density of water is 1.00 g/mL, the mass is 1000.0 grams.

Now, we can calculate the change in temperature using the equation:

q = mcΔT

-97.66 kJ = 1000.0 g * 4.184 J/g-K * ΔT

ΔT = -97.66 kJ / (1000.0 g * 4.184 J/g-K) = -0.0232 K = -0.0232 °C

Since the change in temperature is negative, the final temperature of the water will be 25.0 °C - 0.0232 °C = 24.9768 °C.

Answer 2

Final temperature of water is
`48.3^(0)\textrm{C}`

Step-by-step explanation:

1 mol of LiBr releases 48.83 kJ of heat upon dissolution in water.

So, 2 moles of LiBr release
`(2* 48.83)kJ` or 97.66 kJ of heat upon dissolution in water.

This amount of heat is consumed by 1000.0 g of water. Hence temperature of water will increase.

Let's say final temperature of water is
`t^(0)\textrm{C}`.

So, change in temperature (
`\Delta T`) of water is
`(t-25)^(0)\textrm{C}` or (t-25) K

Heat capacity (C) of water is
`4.184(J)/(g.K)`

Hence,
`m_(water)* C_(water)* \Delta T_(water)=97.66* 10^(3)J`

where m is mass

So,
`(1000.0g)* (4.184(J)/(g.K))* (t-25)K=97.66* 10^(3)J`

or,
`t=48.3`

Hence final temperature of water is
`48.3^(0)\textrm{C}`