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Chapter 14: Problem 9

In the reaction \(A(g) \longrightarrow 2 B(g)+C(g),\) the total pressure increases while the partial pressure of \(\mathrm{A}(\mathrm{g})\) decreases. If the initial pressure of \(\mathrm{A}(\mathrm{g})\) in a vessel of constant volume is \(1.000 \times 10^{3} \mathrm{mmHg}\) (a) What will be the total pressure when the reaction has gone to completion? (b) What will be the total gas pressure when the partial pressure of \(\mathrm{A}(\mathrm{g})\) has fallen to \(8.00 \times 10^{2} \mathrm{mmHg} ?\)

Short Answer

Expert verified
The total pressure when the reaction has gone to completion is \(3.000 \times 10^{3} \mathrm{mmHg}\). The total pressure when the partial pressure of \(A(g)\) has fallen to \(8.00 \times 10^{2} \mathrm{mmHg}\) is \(1.600 \times 10^{3} \mathrm{mmHg}\).

Step by step solution

01

Understand the relationship between reactant and products

In the reaction \(A(g) \longrightarrow 2 B(g)+C(g)\), for every molecule of \(A(g)\) that reacts, it produces two molecules of \(B(g)\) and one molecule of \(C(g)\). Therefore, for every decrease in the pressure of \(A(g)\), the total pressure increases by an equivalent of three times that decrease (This is due to the increase of two molecules of \(B(g)\) plus one molecule of \(C(g)\)).
02

Calculating the total pressure when reaction has gone to completion

If the reaction is complete, all the \(A(g)\) will be converted into \(B(g)\) and \(C(g)\). Therefore the pressure of \(A(g)\) will be 0. Since there was a total reduction of \(1.000 \times 10^{3} \mathrm{mmHg}\) in \(A(g)\), the total pressure increases by three times this which gives a total pressure of \(1.000 \times 10^{3} \times 3 \mathrm{mmHg}\) or \(3.000 \times 10^{3} \mathrm{mmHg}\) after the reaction.
03

Determining the total pressure when partial pressure of \(A(g)\) has dropped to \(8.00 \times 10^{2} mmHg\)

In this case, the decrease in the pressure of \(A(g)\) is \(1.000 \times 10^{3}\) \(mmHg\) - \(8.00 \times 10^{2}\) \(mmHg\) = \(2.00 \times 10^{2} mmHg\). Applying the 1:3 ratio as identified in Step 1, the total pressure will increase by three times this decrease. Therefore, the total pressure in this case is equal to the initial pressure \(1.000 \times 10^{3} mmHg\) plus the increase which is \(2.00 \times 10^{2} \times 3 mmHg\) or \(6.00 \times 10^{2} mmHg\). Adding these together gives a total pressure of \(1.600 \times 10^{3} mmHg\).

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