Given below are two statements :
Statement I : The rate law for the reaction A+B → C is rate (r)=k[A]²[B]. When the concentration of both A and B is doubled, the reaction rate is increased "x" times.
Statement II :
The figure is showing "the variation in concentration against time plot" for a "y" order reaction.
The Value of x+y is __________.

Solution

Chemistry ma’am · 3 steps

Doubling both concentrations multiplies the rate by 8, and a straight [R] vs t line means zero order.

1. Find x. In the rate law each concentration is raised to its order. Doubling [A] multiplies the rate by 2², and doubling [B] multiplies it by 2¹.

   r₂/r₁ = k(2[A])²(2[B])/k[A]²[B] = 2² × 2 = 8

2. Find y. The plot of [R] against time is a straight line starting at [R]₀ with slope -k. That is the integrated zero-order law [R] = [R]₀ - kt, so y = 0.
3. So. x + y = 8 + 0 = 8.

The common slip: 9. Reading the falling straight line as first order sets y = 1 and gives 9. First order is a straight line only for ln[R] against time, not for [R].

The idea to keep. Doubling a concentration scales the rate by 2ⁿ, where n is that reactant's order. For graphs, a straight [R] vs t line is zero order, and a straight ln[R] vs t line is first order.

Answer from the NTA final answer key. Solution by PrepWiser.