Solution Found!
Calculate the work done in joules by the reaction when
Chapter 6, Problem 83P(choose chapter or problem)
Calculate the work done in joules by the reaction
\(2 \mathrm{Na}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{NaOH}(a q)+\mathrm{H}_{2}(g)\)
when \(0.34 \mathrm{~g}\) of Na reacts with water to form hydrogen gas at \(0^{\circ} \mathrm{C}\) and \(1.0\) atm.
Questions & Answers
QUESTION:
Calculate the work done in joules by the reaction
\(2 \mathrm{Na}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{NaOH}(a q)+\mathrm{H}_{2}(g)\)
when \(0.34 \mathrm{~g}\) of Na reacts with water to form hydrogen gas at \(0^{\circ} \mathrm{C}\) and \(1.0\) atm.
ANSWER:
Here, we are going to calculate the work done in joules by the reaction.
Step 1 of 2
We have to calculate the volume of the system.
\(2 \mathrm{Na}(\mathrm{s})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow 2 \mathrm{NaOH}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{~g})\)
Given that,
Mass of \(\mathrm{Na}=0.34 \mathrm{~g}\)
Molar mass \(\mathrm{Na}=23.0 \mathrm{~g} / \mathrm{mol}\)
Therefore, the moles of \(\mathrm{H}_{2}\) would be
\(0.34 \mathrm{~g} \mathrm{Na} \times \frac{1.0 \operatorname{mol} \mathrm{Na}}{23.0 \mathrm{gNa}} \times \frac{1.0 \mathrm{~mol} H_{2}}{2.0 \mathrm{~mol} \mathrm{Na}}=\mathbf{0 . 0 0 7 3 9} \mathrm{mol}\) of \(\mathrm{H}_{2}\)
From the ideal gas equation,
Given that,
Therefore, It is given that
Pressure, \(\mathrm{P}=1.0 \mathrm{~atm}\)
Temperature, \(\mathrm{T}=0^{\circ} \mathrm{C}=0+273=273 \mathrm{~K}\)
Gas constant, \(\mathrm{R}=0.0821\) atm. \(\mathrm{L} / \mathrm{mol} . \mathrm{K}\)
\(\begin{aligned}&\begin{aligned}P V &=n R T \\\Rightarrow V &=\frac{n R T}{P} \\\Rightarrow &-(\mathbf{1}) \\\Rightarrow V &=\frac{0.00739 \text { moles } \times 0.0821 \mathrm{~atm} . L / \text { mol. } K \times 273 K}{1.0 \mathrm{~atm}} \\&=0.166 \mathrm{~L}\end{aligned} \end{aligned}\)
Therefore, change in volume of the system \(=0.166 \mathrm{~L}\)