Question

Find the enthalpy change per mole of sodium when sodium reacts with water. 8 grams of sodium reacts with 227 cm3 of water, producing a temperature change from 298 K to 308.4 K. The specific heat capacity of water is 4.18 J/K g. A. -28.36 kJ B. 9868.1 J C. 62.75 kJ D. -28356.7 kJ

Answer 1

The correct answer to the following question will be Option A (-28.36 KJ). The further explanation is given below.

Step-by-step explanation:

The given values are:

Sodium's mass = 8 grams

Water's volume = 227 cm³

Molar mass = 23 g/mol

Density of water = 1 g/cm³

Specific heat capacity (
`C_(w)`) = 4.18 J/Kg

As we know,

⇒
`Moles \ of \ Na =(mass \ of \ Na)/(Molar \ mass \ of \ Na)`

On putting the estimated values, we get

⇒
`=(8)/(23)`

⇒
`=0.348 \ mol`

Now, Water's mass will be:

⇒
`M_(w) = density* volume`

`=1* 227`

`=227 \ g`

Change in temperature will be:

⇒
`\Delta T=(308.4-298)K`

`=10.4 \ K`

Heat released will be:

⇒
`q=M_(w)* C_(w)* \Delta T`

On substituting the estimated values, we get

`=227* 4.18* 10.4`

`=9.87* 10^3`

`=9.87 \ KJ`

So that the change in the solution of Na will be:

⇒
`\Delta H=(-q)/(Moles \ of \ Na)`

`=(-9.87)/(0.348)`

`=-28.36 \ KJ/mol`