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I am given a table that gives me the different total pressures at certain times. The decomposition of phosphine is the reaction:

$$\ce{4 PH3(g) <=> P4(g) + 6 H2(g)}$$

The time/pressure values are:

\begin{array}{rr} \mathrm{time\ [min]} & \mathrm{pressure\ [Torr]} \\ \hline 0 & 100 \\ 40 & 150\\ 80 & 167\\ 100 & 172\\ \end{array}

This reaction takes place at 950K. I can then find the molarity of the total gas mixture at each time using $$\frac{n}{V}=\frac{P}{RT}$$

I need to know how to get the pressure of the $\ce{P4}$ for each individual time. How would one go about doing this?

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The assumption you need to make is that at time zero, only $\ce{PH3}$ is present. So relative to the moles at time zero, or $N_0$, you should find $\Delta N = N(t) - N_0$, the total new moles formed by the reaction since it began.

The reaction converts 4 moles of gas to 7.

So pressure increases mean the reaction is moving to the right. Every mole of pressure increase must be $\frac{6}{7}$ due to $\ce{H_2}$ and $\frac{1}{7}$ due to $\ce{P4}$ formation.

So just find the total new moles of gas and multiply by $\frac{1}{7}$.

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  • $\begingroup$ So for time equal to 0 the concentration of P4 is 0 as well, correct? $\endgroup$ – user3138766 Mar 2 '15 at 17:29
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    $\begingroup$ That's the assumption I would make. $\endgroup$ – Curt F. Mar 2 '15 at 21:48

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