Question

Sodium metal reacts with water to produce hydrogen gas and sodium hydroxide according to the chemical equation shown below. When 0.030 mol of Na is added to 100.00g of water, the temperature of the resulting solution rises from 25.00°C to 37.90°C. If the specific heat of the solution is 4.18 J/(g · °C), calculate ?H for thereaction, as written.
2 Na(s) + 2 H2O(l) --> 2 NaOH(aq) + H2(g)
?H= ?
Answer is one of the following:
-90.0 kJ
-360 kJ
-5.41 kJ
-180 kJ

Answer 1

The answer is: ΔH = -362 kJ. See solution below:

Mass of Na = moles x molar mass

= 0.030 x 22.99 = 0.6897 g

Mass of solution = mass of water + mass of Na

= 100 + 0.6897 = 100.6897

Heat released by reaction of 0.030 mol of Na = heat absorbed by solution

= mass x specific heat x temperature change of solution

= 100.6897 x 4.18 x (37.90 - 25.00)

= 5429.39 J

Heat released by reaction of 2 mol of Na = 2/0.030 x 5429.39

= 3.62 x 10⁵ J

ΔH = -Heat released by reaction of 2 mol of Na

= -3.62 x 10⁵ J = -362 kJ

Note that ΔH is negative as reaction is exothermic and heat is released

Answer 2

The total heat released is Q=cmΔT;

(c is the specific heat of water, m is the mass of water, ΔT is the change in temperature.)

So Q=4.18J/(g*C^o)*100.00g*(37.90C^o-25.00C^o)=5392.2J=5.3922kJ

in this reaction, 2Na(s) + 2H₂O(l) ---2NaOH (aq) + H₂ (g)

0.030mol Na reacted with water; so there's 0.030/2=0.015mol of reaction going on.

The enthalpy change of the reaction is just the heat of reaction per mole; so the absolute value of ΔH=Q/mols of reaction=5.3922kJ/0.015mol=359.48kJ/mol

because this is an exothermic reaction, the ΔH<0; so ΔH=-359.48kJ/mol

Answer 3

Answer:

${\Delta }_{\text{r}}H={\text{-181 kJ⋅mol}}^{\text{-1}}$

Explanation:

There are two heat transfers to consider:

$\text{heat released by reaction + heat absorbed by water = 0}$
$mmmmmm{q}_{1}mmmmml+mmmmmm{q}_{2}mmmml=0$
$mmmmmn{\Delta }_{\text{r}}Hmmmml+mmmmlm{C}_{\text{s}}\Delta Tmmm=0$

Let's calculate each of these heats separately.

Step 1. Calculate $q_1$

$\text{Moles of Na}=0.575\text{g Na}\times \frac{\text{1 mol Na}}{22.99\text{g Na}}=\text{0.025 01 mol Na}$

$q_1=n{\Delta }_{\text{r}}H=\text{0.025 01}{\Delta }_{\text{r}}Hl\text{mol}$

Step 2. Calculate $q_2$

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

$ml=\text{100.00 g + 0.575 g = 100.575 g}$
$C_{\text{s}}l={\text{4.184 J⋅°C}}^{\text{-1}}{\text{g}}^{\text{-1}}$
$\Delta T=\text{35.75 °C - 25.00 °C = 10.75 °C}$

∴ $q_2=100.575\text{g}\times \text{4.184 J}\cdot {\text{°C}}^{\text{-1}}{\text{g}}^{\text{-1}}\times 10.75\text{°C}=\text{4523.7 J}$

Step 3. Calculate ${\Delta }_{\text{r}}H$

$q_1+q_2=0$

$\text{0.025 01}{\Delta }_{\text{r}}Hl\text{mol}+\text{4523.7 J}=0$

${\Delta }_{\text{r}}H=\frac{\text{-4523.7 J}}{\text{0.025 01 mol}}={\text{-181 000 J⋅mol}}^{\text{-1}}={\text{-181 kJ⋅mol}}^{\text{-1}}$