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A reaction $\ce{A + 2 B -> C}$ is carried out in constant volume at $\pu{227 ^\circ C}$. the volume is $\pu{2 L}$. There is no $\ce{C}$ at the beginning. $[\ce{A}]_0 = \pu{0.035 mol L-1}$. $\ce{B}$ concentration changes over time.

\begin{array}{l|cccccccc} t~(\pu{min})& 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline [\ce{B}]~(\pu{mol L-1}) & 0.09 & 0.066 & 0.05 & 0.039 & 0.033 & 0.028 & 0.024& 0.023 \end{array}

a) At $t = 5$, what is the conversion rate?
b) Calculate the reaction rate of $\ce{C}$ by using graph method according to $\ce{C}$ at time $0$ and $6$.
c) Calculate the total pressure at time $3$.
d) If this reaction should have done at constant pressure what would be the volume change rate?

I answered the a, c and d questions, but I am not sure about it. I can not solve question b, graph method, these were my exam questions and I want to know correct answers. Here are my answers:


Part a.

$$t = \pu{5 min}$$ $$\ce{\underset{0.035}{A (g)} + 2 \underset{0.09}{B (g)} -> C (g)}$$

$\ce{A}$ is a limiting reactant. At constant volume

\begin{align} [\ce{A}] &= [\ce{A}]_0 (1 - x) \\ [\ce{B}] &= [\ce{B}]_0 - 2x[\ce{A}]_0 \end{align}

$$\pu{0.028 mol L-1} = \pu{0.09 mol L-1} - 2 \cdot \pu{0.035 mol L-1} \cdot x \to x = 0.886$$


Part b.

$$-\ln{(1 - x)} = kt$$ $$t = 6$$ $$0.024 = 0.09 - 2 \cdot 0.035 \cdot x \to x = 0.943$$ $$-\ln{(1 - 0.943)} = 6k \qquad \color{red}{?}$$


Part c.

$$p_T = C_T RT$$ $$0.039 = 0.09 - 2 \cdot 0.035 \cdot x \to x = 0.728$$ \begin{align} [\ce{A}] &= 0.035 (1 - 0.728) = 9.52 \cdot 10^{-3} \\ [\ce{B}] &= 0.039 \\ [\ce{C}] &= [\ce{A}]_0 \cdot x = 0.035 \cdot 0.728 \end{align} $$C_T = 0.074$$ $$p_T = \pu{0.076 mol L-1} \cdot \pu{0.082 L atm mol-1 K-1} \cdot \pu{(273 + 227) K} = \pu{3.034 atm}$$


Part d.

$$\epsilon = \frac{1 - 3}{3} = -\frac{2}{3}$$ $$V = V_0(1 + \epsilon x) \qquad \color{red}{?}$$

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    $\begingroup$ I edited your question using MathJax and Markdown according to the guidelines (more or less). Please feel free to check the correctness and improve further (see formatting tips here) $\endgroup$ – andselisk Dec 4 '17 at 15:48
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    $\begingroup$ @andselisk I'm not sure if you putting a ton of effort into formatting everybody's questions is a flaw... (if you nominate yourself for the next Mod elections, whenever it is, you'll have my vote :D) $\endgroup$ – paracetamol Dec 4 '17 at 15:53

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