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Question: Calculate the work done (in J) when 4.5 g \({{H}_{2}}{{O}_{2}}\) reacts against a pressure of 1 atm ...

Calculate the work done (in J) when 4.5 g H2O2{{H}_{2}}{{O}_{2}} reacts against a pressure of 1 atm at 25∘C{{25}^{\circ }}C.
2H2O2(l)→O2(g)+H2O(l)2{{H}_{2}}{{O}_{2}}(l)\to {{O}_{2}}(g)+{{H}_{2}}O(l)
(a) −1.49 x 102-1.49\ \text{x 1}{{\text{0}}^{2}}
(b) 4.5 x 1024.5\ \text{x 1}{{\text{0}}^{2}}
(c) 3.2 x 1023.2\ \text{x 1}{{\text{0}}^{2}}
(d) −6.1 x 102-6.1\ \text{x 1}{{\text{0}}^{2}}

Explanation

Solution

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 H2O2{{H}_{2}}{{O}_{2}}is 4.5 g
The molecular mass of H2O2{{H}_{2}}{{O}_{2}}is 34 g
So, the number of moles of H2O2{{H}_{2}}{{O}_{2}}, will be
Moles = Given massMolecular mass = 4.534 = 0.132 mol\text{Moles = }\dfrac{\text{Given mass}}{\text{Molecular mass}}\text{ = }\dfrac{4.5}{34}\text{ = 0}\text{.132 mol}

So, the given reaction is:
2H2O2(l)→O2(g)+H2O(l)2{{H}_{2}}{{O}_{2}}(l)\to {{O}_{2}}(g)+{{H}_{2}}O(l)
According to the reaction, 2 moles of H2O2{{H}_{2}}{{O}_{2}} produces 1 mole of oxygen. So, when 0.132 mole of H2O2{{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.
Δng = np−nr\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:
Δng = np−nr=(0.066+0)−0=0.066\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, ΔV=Δng x 22.4 = 0.066 x 22.4 = 1.48\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=−PΔVW=-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=−PΔV=−(1 x 1.48) = -1.48 L-atmW=-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 x 102J-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.