Question

The following data were measured for the reaction BF3(g)+NH3(g)→F3BNH3(g): Experiment [BF3](M) [NH3](M) Initial Rate (M/s) 1 0.250 0.250 0.2130 2 0.250 0.125 0.1065 3 0.200 0.100 0.0682 4 0.350 0.100 0.1193 5 0.175 0.100 0.0596 Part A What is the rate law for the reaction? What is the rate law for the reaction? rate=k[BF3]2[NH3] rate=k[BF3][NH3] rate=k[BF3][NH3]2 rate=k[BF3]2[NH3]2

Answer 1

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

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

Resolving  

$\alpha=1$

This means that the rate law for this reaction is  

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