Determine the partial pressure of A and B.

Recall that the partial pressure of a gas (P_{Gas}) in a mixture is given by:

$\overline{){{\mathbf{P}}}_{{\mathbf{gas}}}{\mathbf{=}}{{\mathbf{\chi}}}_{{\mathbf{gas}}}{{\mathbf{P}}}_{{\mathbf{total}}}}$

where:

χ_{Gas} = mole fraction of the gas

P_{total} = total pressure of the gas mixture

Assuming the volume fraction is equal to the mole fraction.

${\mathbf{\chi}}_{\mathbf{A}}\mathbf{=}\frac{\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{}\mathbf{L}}{\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{}\mathbf{L}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{1}\mathbf{}\mathbf{L}}\phantom{\rule{0ex}{0ex}}{\mathbf{\chi}}_{\mathbf{A}}\mathbf{=}\frac{\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{}\overline{)\mathbf{L}}}{\mathbf{3}\mathbf{.}\mathbf{7}\mathbf{}\overline{)\mathbf{L}}}$

**χ _{A} = 0.4324**

${\mathbf{\chi}}_{\mathbf{B}}\mathbf{=}\frac{\mathbf{2}\mathbf{.}\mathbf{1}\mathbf{}\mathbf{L}}{\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{}\mathbf{L}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{1}\mathbf{}\mathbf{L}}\phantom{\rule{0ex}{0ex}}{\mathbf{\chi}}_{\mathbf{B}}\mathbf{=}\frac{\mathbf{2}\mathbf{.}\mathbf{1}\mathbf{}\overline{)\mathbf{L}}}{\mathbf{3}\mathbf{.}\mathbf{7}\mathbf{}\overline{)\mathbf{L}}}$

**χ _{B} = 0.5675**

Determine the partial pressure of A and B:

The following reaction occurs in a closed container:

A(g) + 2 B(g) → 2 C(g)

A reaction mixture initially contains 1.6 L of A and 2.1 L of B.

Assuming that the volume and temperature of the reaction mixture remain constant, what is the percent change in pressure if the reaction goes to completion?

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