____________ In a group all on their own. No neutrons.

1 answer:

6 0

These are they only ones you need? Really it looks like the whole test. Also only 5 points for all this work I’m sorry no thank you

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The following data were measured for the reaction BF3(g)+NH3(g)→F3BNH3(g): Experiment [BF3](M) [NH3](M) Initial Rate (M/s) 1 0.2

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Answer:

$-r_{A}=k\times[BF_3]^{1}\times[NH_3]^{1}$

Explanation:

The rate law of a chemical reaction is given by

$-r_{A}=k\times[BF_3]^{\alpha}\times[NH_3]^{\beta}$

This law can be written for any experiment, and making the quotient between those expressions the reaction orders can be found

Between experiments 1 and 2

$\frac{-r_{A1}}{{-r}_{A2}}=\left(\frac{\left[NH_3\right]_1}{\left[NH_3\right]_2}\right)^\beta$

Then the expression for the calculation of $\beta$

$\beta=\frac{ln\frac{-r_{A1}}{-r_{A2}}}{ln\left(\frac{\left[NH_3\right]_1}{\left[NH_3\right]_2}\right)}=\frac{ln\frac{0.2130}{0.1065}}{ln\left(\frac{0.250}{0.125}\right)}$

Resolving

$\beta=1$

Doing the same between experiments 3 and 4 the expression for $\alpha$ is

$\alpha=\frac{ln\frac{-r_{A3}}{-r_{A4}}}{ln\left(\frac{\left[BF_3\right]_3}{\left[BF_3\right]_4}\right)}=\frac{ln\frac{0.0682}{0.1193}}{ln\left(\frac{0.200}{0.350}\right)}$