Question

Sodium metal reacts with water to produce hydrogen gas and sodium hydroxide. When 0.575g of Na is added to 100.00g of water, the temperature of the water rises from 25.00°C to 35.75°C. Calculate ΔH for the reaction, in kJ/mol. How do I solve this?

This involves math

Answer 1

`Δ_text(r)H = "-181 kJ·mol"^"-1"`

Explanation:

There are two heat transfers to consider:

`"heat released by reaction + heat absorbed by water = 0"`
`color(white)(mmmmmm)q_1 color(white)(mmmmml)+color(white)(mmmmmm) q_2 color(white)(mmmml)= 0`
`color(white)(mmmmm)nΔ _text(r)H color(white)(mmmml)+ color(white)(mmmml)mC_text(s)ΔT color(white)(mmm)= 0`

Let's calculate each of these heats separately.

Step 1. Calculate `q_1`

#"Moles of Na" = 0.575 color(red)(cancel(color(black)("g Na"))) × "1 mol Na"/(22.99 color(red)(cancel(color(black)("g Na")))) = "0.025 01 mol Na"#

`q_1 = nΔ_text(r)H = "0.025 01"Δ_text(r)Hcolor(white)(l) "mol"`

Step 2. Calculate `q_2`

I assume that the specific heat capacity of the solution is the same as for water. Then

`m color(white)(l)= "100.00 g + 0.575 g = 100.575 g"`
`C_text(s) color(white)(l)= "4.184 J·°C"^"-1""g"^"-1"`
`ΔT = "35.75 °C - 25.00 °C = 10.75 °C"`

∴ `q_2 = 100.575 color(red)(cancel(color(black)("g"))) × "4.184 J"·color(red)(cancel(color(black)("°C"^"-1""g"^"-1"))) × 10.75 color(red)(cancel(color(black)("°C"))) = "4523.7 J"`

Step 3. Calculate `Δ_text(r)H`

`q_1 + q_2 = 0`

`"0.025 01"Δ_text(r)Hcolor(white)(l) "mol" + "4523.7 J" = 0`

`Δ_text(r)H = "-4523.7 J"/"0.025 01 mol" = "-181 000 J·mol"^"-1" = "-181 kJ·mol"^"-1"`