This problem involves determining the order of a reaction from concentration vs. time data. The reaction is R → P, and we are given the concentration of R at different times.

Step 1: Understand Reaction Order
The order of a reaction defines how the rate depends on the concentration of reactants. For a reaction R → P, the rate law is: $r=-\frac{d\left[R\right]}{dt}=k{\left[R\right]}^n$ where n is the order.

Step 2: Analyze the Data
We have:

[R] (M)	1.0	0.75	0.40	0.10
t (min)	0.0	0.05	0.12	0.18

Notice that the concentration decreases very rapidly in a short time. For example, from t=0 to t=0.05 min, [R] drops from 1.0 M to 0.75 M. This suggests a high-order reaction.

Step 3: Test for Zero Order
For zero order (n=0), [R] should decrease linearly with time: $\left[R\right]=[R{]}_0-kt$. Calculate the rate of decrease between intervals:

• From t=0 to t=0.05: Δ[R]/Δt = (0.75-1.0)/(0.05-0) = -0.25/0.05 = -5 M/min
• From t=0.05 to t=0.12: Δ[R]/Δt = (0.40-0.75)/(0.12-0.05) = -0.35/0.07 = -5 M/min
• From t=0.12 to t=0.18: Δ[R]/Δt = (0.10-0.40)/(0.18-0.12) = -0.30/0.06 = -5 M/min

The rate is constant at -5 M/min, which is characteristic of a zero-order reaction.

Step 4: Verify with Integrated Rate Law
The integrated rate law for zero order is: $\left[R\right]=[R{]}_0-kt$. Using k=5 M/min and [R]₀=1.0 M:

• At t=0.05: [R] = 1.0 - (5)(0.05) = 0.75 M ✓
• At t=0.12: [R] = 1.0 - (5)(0.12) = 0.40 M ✓
• At t=0.18: [R] = 1.0 - (5)(0.18) = 0.10 M ✓

The calculated values match the given data perfectly.

Final Answer: The reaction is zero order.

Related Topics & Formulae

Zero-Order Reaction:
- Rate is independent of reactant concentration.
- Rate law: $r=k$
- Integrated law: $\left[R\right]=[R{]}_0-kt$
- Half-life: $t_{1/2}=\frac{[R{]}_0}{2k}$

Other Common Orders:
- First order: Rate ∝ [R]; linear plot of ln[R] vs. t.
- Second order: Rate ∝ [R]²; linear plot of 1/[R] vs. t.