**Part A.**

We are asked to determine the **Rate Law **that describes the decomposition of Phosphine, given the temperature and the total system pressure as a function of time.

The chemical equation is:

4PH_{3(g)} → P_{4(g)} + 6H_{2(g)}

Recall that the rate of reaction is the change in pressure (in this case) of a reactant for every change in time. Since we are given with the *Total pressure* of the system and we are only concerned with the *reactant*, we need to first identify the Pressure of PH_{3 }for every given time.

At t=0, there was only PH_{3}, therefore:

${P}_{Total}={P}_{P{H}_{3}}=$**100 torr**^{ }

At t=40 min, we need to use the ICF chart to determine the pressure of PH_{3}.

4PH_{3(g)} → P_{4(g)} + 6H_{2(g)}

At t=40 min, total system pressure is 151 torr, therefore,

$100-4x+x+6x=151\phantom{\rule{0ex}{0ex}}100+3x=151\phantom{\rule{0ex}{0ex}}3x=151-100\phantom{\rule{0ex}{0ex}}\frac{\overline{)3}x}{\overline{)3}}=\frac{51}{3}$

x = **17**

Therefore, pressure of PH_{3 }at t=40 is:

${P}_{P{H}_{3}}=100torr-\left(4x\right)torr\phantom{\rule{0ex}{0ex}}{P}_{P{H}_{3}}=100torr-\left(4\left(17\right)\right)torr\phantom{\rule{0ex}{0ex}}{P}_{P{H}_{3}}=32torr$

**At t=40 min, pressure of PH _{3} is**

At t=80 min, total system pressure is 168 torr, therefore,

$100-4x+x+6x=168\phantom{\rule{0ex}{0ex}}100+3x=168\phantom{\rule{0ex}{0ex}}3x=168-100\phantom{\rule{0ex}{0ex}}\frac{\overline{)3}x}{\overline{)3}}=\frac{68}{3}$

* x = **22.7*

Therefore, pressure of PH_{3 }at t=80 is:

${P}_{P{H}_{3}}=100torr-\left(4x\right)torr\phantom{\rule{0ex}{0ex}}{P}_{P{H}_{3}}=100torr-\left(4(22.7)\right)torr\phantom{\rule{0ex}{0ex}}{P}_{P{H}_{3}}=9.2torr$

**At t=80 min, pressure of PH _{3} is**

At t=100 min, total system pressure is 171 torr, therefore,

$100-4x+x+6x=171\phantom{\rule{0ex}{0ex}}100+3x=171\phantom{\rule{0ex}{0ex}}3x=171-100\phantom{\rule{0ex}{0ex}}\frac{\overline{)3}x}{\overline{)3}}=\frac{71}{3}$

* x = **23.7*

Therefore, pressure of PH_{3 }at t=100 is:

${P}_{P{H}_{3}}=100torr-\left(4x\right)torr\phantom{\rule{0ex}{0ex}}{P}_{P{H}_{3}}=100torr-\left(4(23.7)\right)torr\phantom{\rule{0ex}{0ex}}{P}_{P{H}_{3}}=5.2torr$

**At t=100 min, pressure of PH _{3} is**

Phosphine, PH_{3}(g), decomposes according to the equation

4PH_{3}(g) → P_{4}(g) + 6H_{2 }(g)

The kinetics of the decomposition of phosphine at 950 K was followed by measuring the total pressure in the system as a function of time. The data to the right were obtained in a run where the reaction chamber contained only pure phosphine at the start of the reaction. Choose the rate law that describes this reaction.

Calculate the value of the rate constant and choose the correct units.

k =

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