Question

Determine the rate law, including the values of the orders and rate law constant, for the following reaction using the experimental data provided.
A + B yields products
Trial [A] [B] Rate
1 0.15 M 0.12 M 1.3 × 10-2 M/min
2 0.30 M 0.12 M 2.6 × 10-2 M/min
3 0.15 M 0.24 M 5.2 × 10-2 M/min
I'm behind in school currently and I'm trying to complete assignments asap. Please help.

Answer 1

well, what you can do is this: to determine the order of the reaction with respect to either x and y, you must choose two reactions where one of the concentrations is constant say trials 1 and 3.

in trials 1 and 3 the concentration of A is constant so we can omit that.

\[(\frac{ 0.15 }{ 0.15 })[A]^{x} *[B]^{y} (\frac{ 0.12 }{ 0.24}) k = (\frac{ 1.3 }{ 5.2})\]

\[[\frac{ 1 }{ 2 }]^{y}* = \frac{ 1 }{ 4 }\]

\[y*\log[\frac{ 1 }{ 2 }] = \log[\frac{ 1 }{ 4 }]\]

\[\frac{ \log[\frac{ 1 }{ 4 } ]}{ \log[\frac{ 1 }{ 2 }] } = 2\]

we can actually do the same thing for A by choosing two reactions where B is constant.

\[([\frac{ 0.15 }{ 0.30 })]^{x} *[(\frac{ 0.12 }{ 0.12})]^{2} k = (\frac{ 1.3 }{ 2.6})\]


\[[\frac{ 1 }{ 2 }]^{x}* = \frac{ 1 }{ 2 }\]


\[x*\log[\frac{ 1 }{ 2 }] = \log[\frac{ 1 }{ 2 }]\]

so clearly x = 1 and y = 2

Therefore, the reaction is first order with respect to A and second order with respect to B

\[[A]^{1}[B]^{2}k = r \]

the rate constnat k is a constant. so you can easily figure that out, by taking the rate and concentrations of A and B for any reaction trial.

I'll leave that part up to you.

\[k = (\frac{ r }{ [A]^{1}[B]^{2} })\]