A rigid container holds an equal number of moles of $N_{2}$ and $H_{2}$ gas at a total pressure of 10.0 atm. The
gases react according to the equation,
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 {NH}_{3} (g)$
If the total pressure of the gas decreases at a rate of
$0.20 atm \cdot s^{- 1}$ , what is the rate of change of the partial pressure of $N_{2}$ in the container?

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Decreases at $0.20 atm \cdot s^{- 1}$

Decreases at $0.10 atm \cdot s^{- 1}$

Decreases at $0.40 atm \cdot s^{- 1}$

Decreases at $0.30 atm \cdot s^{- 1}$

answer is D.

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Detailed Solution

In the reaction we have,

$N_{2} + 3 H_{2} ⇌ 2 {NH}_{3}$

$\begin{array}{ccc} 5 - x & 5 - 3 x & 2 x \end{array}$

Thus,

$P_{Total} = 10 - 2 x$

$\frac{d P_{Total}}{d t} = - 2 \frac{d x}{d t} = 0.2$

$\frac{d N_{2}}{d t} = - \frac{d x}{d t} = 0.1$

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