Question

Consider the reaction: 2A(g)+B(g)→3C(g).

Answer 1

Answer:

Part A

$Rate = -\frac{1}{2}\frac{\Delta A}{\Delta t} =-\frac{\Delta B}{\Delta t} = \frac{1}{3}\frac{\Delta C}{\Delta t}$

Part B

$-\frac{\Delta B}{\Delta t}= 0.0500 M s^{-1}$

Part C

$\frac{\Delta C}{\Delta t} = 0.15 M s^{-1}$

Explanation:

For a general reaction,

$aA(g) + bB(g) \rightarrow cC(g)$

Rate is given by:

Rate: $Rate = -\frac{1}{a}\frac{\Delta A}{\Delta t} =-\frac{1}{b}\frac{\Delta B}{\Delta t} = \frac{1}{c}\frac{\Delta C}{\Delta t}$

So, for the given reaction:

$2A(g) + B(g) \rightarrow 2C(g)$

$Rate = -\frac{1}{2}\frac{\Delta A}{\Delta t} =-\frac{\Delta B}{\Delta t} = \frac{1}{3}\frac{\Delta C}{\Delta t}$

Part B

$-\frac{1}{2}\frac{\Delta A}{\Delta t} =-\frac{\Delta B}{\Delta t}$

Given: $-\frac{\Delta A}{\Delta t} = 0.100\ Ms^{-1}$

$\frac{1}{2}\frac{0.100}{\Delta t} =-\frac{\Delta B}{\Delta t}$

$-\frac{\Delta B}{\Delta t}$ = 0.0500 M s^-1

Part C

$-\frac{\Delta B}{\Delta t} =\frac{1}{3}\frac{\Delta C}{\Delta t}$

$-\frac{\Delta B}{\Delta t} =\frac{1}{3}\frac{\Delta C}{\Delta t}$

$0.0500 = \frac{1}{3}\frac{\Delta C}{\Delta t}$

$\frac{\Delta C}{\Delta t} = 3 \times 0.0500 = 0.15 M s^{-1}$