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

Answer:

Final temperature of water is $48.3^{0}\textrm{C}$

Explanation:

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

So, 2 moles of LiBr release $(2\times 48.83)kJ$ 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 $4.184\frac{J}{g.K}$

Hence, $m_{water}\times C_{water}\times \Delta T_{water}=97.66\times 10^{3}J$

where m is mass

So, $(1000.0g)\times (4.184\frac{J}{g.K})\times (t-25)K=97.66\times 10^{3}J$

or, $t=48.3$

Hence final temperature of water is $48.3^{0}\textrm{C}$