Question

For a chemical reaction \[\ce{2A + B -> 2C + 3D}\], the rate of disappearance of A is 0.10 mol L⁻¹ s⁻¹. Calculate the rate of reaction and the rate of appearance of D.

Answer 1

The given reaction is \[\ce{2A + B -> 2C + 3D}\] and

the rate of disappearance of A is

\[\ce-{\frac{d[A]}{dt}}\]

= 0.10 mol L⁻¹ s⁻¹

From the stoichiometry

Rate of reaction = \[\ce{\frac{1}{2}-\frac{d[A]}{dt}}\]

= \[\ce{\frac{1}{2} \times 0.10}\]

Rate of reaction = 0.05 mol L⁻¹ s⁻¹

From the balanced equation

Rate of appearance of D = \[\ce{\frac{3}{2}- \frac{d[A]}{dt}}\]

= \[\ce{\frac{3}{2} \times 0.10}\]

Rate of appearance of D = 0.15 mol L⁻¹ s⁻¹