Question

A student sets up two reactions. Reaction 1 uses 0.300 mol/L of reactant, and reaction 2 uses 0.600 mol/L of reactant. How many times faster is reaction 2 compared to reaction 1?

Answer 1

Rate reaction 2 is faster in 3 significant figure 2.00

Step-by-step explanation:

Reactant of reaction 1 = 0.3mol/L

Reactant of reaction 2 = 0.6mol/L

To determine how many times faster, we find the ration of reactant 2 to reactant 1

`\begin{gathered} \frac{reac\tan t\text{ 2}}{\text{reactant 1}}\text{ = }(0.6)/(0.3) \\ \frac{reac\tan t\text{ 2}}{\text{reactant 1}}\text{ = }(6)/(3)\text{ = 2} \\ reac\tan t\text{ 2 = }2*\text{ reactant 1} \end{gathered}`

Therefore, reactant 2 is 2 times faster than reactant 1

Rate reaction 2 is faster in 3 significant figures 2.00