Question

7) For the reaction 9A (g) + B (g)  5C(g) + 1/6 D (g), it takes 4 and a half minutes for the concentration of C to increase to 1.33 M. What would be the decrease in A during this time interval? a. 0.27 b. 0.044 c. 0.53

Answer 1

Answer: The correct option is, (C) 0.53

Explanation:

The given chemical reaction is:

$9A(g)+B(g)\rightarrow 5C(g)+\frac{1}{6}D(g)$

The rate of the reaction for disappearance of A and formation of C is given as:

$\text{Rate of disappearance of }A=-\frac{1}{9}\times \frac{\Delta [A]}{\Delta t}$

Or,

$\text{Rate of formation of }C=+\frac{1}{5}\times \frac{\Delta [C]}{\Delta t}$

where,

$\Delta C$ = change in concentration of C = 1.33 M

$\Delta t$ = change in time = 4.5 min

Putting values in above equation, we get:

$\frac{1}{9}\times \frac{\Delta [A]}{\Delta t}=\frac{1}{5}\times \frac{\Delta [C]}{\Delta t}$

$\frac{\Delta [A]}{\Delta t}=\frac{9}{5}\times \frac{\Delta [C]}{\Delta t}$

$\frac{\Delta [A]}{\Delta t}=\frac{9}{5}\times \frac{1.33M}{4.5min}$

$\frac{\Delta [A]}{\Delta t}=0.53M/min$

Thus, the decrease in A during this time interval is, 0.53