Question

Calculate the work done (in J) when 4.5 g ${{H}_{2}}{{O}_{2}}$ reacts against a pressure of 1 atm at ${{25}^{\circ }}C$.
$2{{H}_{2}}{{O}_{2}}(l)\to {{O}_{2}}(g)+{{H}_{2}}O(l)$
(a) $-1.49\ \text{x 1}{{\text{0}}^{2}}$
(b) $4.5\ \text{x 1}{{\text{0}}^{2}}$
(c) $3.2\ \text{x 1}{{\text{0}}^{2}}$
(d) $-6.1\ \text{x 1}{{\text{0}}^{2}}$

Answer 1

Hint: The work done by the system is calculated by multiplying the pressure of the system to the change in volume of the system. Since, the work done by the system, the sign of the work done is taken negatively.

Complete step by step answer:
The number of moles is calculated by dividing the given mass to the molecular mass of the compound.
The given mass of ${{H}_{2}}{{O}_{2}}$is 4.5 g
The molecular mass of ${{H}_{2}}{{O}_{2}}$is 34 g
So, the number of moles of ${{H}_{2}}{{O}_{2}}$, will be
$\text{Moles = }\dfrac{\text{Given mass}}{\text{Molecular mass}}\text{ = }\dfrac{4.5}{34}\text{ = 0}\text{.132 mol}$

So, the given reaction is:
$2{{H}_{2}}{{O}_{2}}(l)\to {{O}_{2}}(g)+{{H}_{2}}O(l)$
According to the reaction, 2 moles of ${{H}_{2}}{{O}_{2}}$ produces 1 mole of oxygen. So, when 0.132 mole of ${{H}_{2}}{{O}_{2}}$ will produce 0.066 mole of oxygen.

For calculating the change in the number of moles will be a difference in the number of moles of product and number of mole reactants.
$\Delta {{n}_{g}}\text{ = }{{n}_{p}}-{{n}_{r}}$
 Only the number of moles of gases is taken.
Hence, the change in the number of moles for the above reaction will be:
 $\Delta {{n}_{g}}\text{ = }{{n}_{p}}-{{n}_{r}}=(0.066+0)-0=0.066$
The change in volume of the gas can be calculated by multiplying the number of moles to the volume of one mole of gas. The standard volume of 1 mole of gas is 22.4 L
Therefore, $\Delta V=\Delta {{n}_{g}}\,\text{x 22}\text{.4 = 0}\text{.066 x 22}\text{.4 = 1}\text{.48}$
The work done by the system is equal to the product of the pressure of the system and the change in volume of the system.

$W=-P\Delta V$
The pressure of the system is 1 atm (given)
And the change in volume is 1.48.

So, the work done is,
$W=-P\Delta V=-(1\text{ x 1}\text{.48) = -1}\text{.48 L-atm}$
Work done is -1.48 L-atm which is equal to $-1.49\ \text{x 1}{{\text{0}}^{2}}J$
So, the correct answer is “Option A”.

Note: According to the sign conventions, w is taken positive if the work is done in the system and w is negative if the work is done by the system. When the work is done in a liter atmosphere it is converted into joules by multiplying it with 101.3 J.