Answer :
Answer: The correct option is, (C) 0.53
Explanation:
The given chemical reaction is:
[tex]9A(g)+B(g)\rightarrow 5C(g)+\frac{1}{6}D(g)[/tex]
The rate of the reaction for disappearance of A and formation of C is given as:
[tex]\text{Rate of disappearance of }A=-\frac{1}{9}\times \frac{\Delta [A]}{\Delta t}[/tex]
Or,
[tex]\text{Rate of formation of }C=+\frac{1}{5}\times \frac{\Delta [C]}{\Delta t}[/tex]
where,
[tex]\Delta C[/tex] = change in concentration of C = 1.33 M
[tex]\Delta t[/tex] = change in time = 4.5 min
Putting values in above equation, we get:
[tex]\frac{1}{9}\times \frac{\Delta [A]}{\Delta t}=\frac{1}{5}\times \frac{\Delta [C]}{\Delta t}[/tex]
[tex]\frac{\Delta [A]}{\Delta t}=\frac{9}{5}\times \frac{\Delta [C]}{\Delta t}[/tex]
[tex]\frac{\Delta [A]}{\Delta t}=\frac{9}{5}\times \frac{1.33M}{4.5min}[/tex]
[tex]\frac{\Delta [A]}{\Delta t}=0.53M/min[/tex]
Thus, the decrease in A during this time interval is, 0.53