Question

Use Figure 18.3 to estimate graphically the instantaneous rate of production of NO at t = 100 s.

Short answer

The instantaneous rate at 100s is \(5.3\; \times {10^{ - 5}}\;mol\;{L^{ - 1}}\)

Step-by-step solution

STEP-1:  Consider the reaction:

The expression for the rate of reaction is for above equation can be written as

\(rate = \;\dfrac{{Change\;in\;concendration}}{{Change\;in\;time}}\)

The rate written with respect to concendration of NO can be given as below expression

\(rate\; = \;\dfrac{{d\left( {NO} \right)}}{{dt}}\)

 The graph between the concendration of NO and  Time in saeconds are as follows:

STEP-3: Finding the  rate of the reaction using graph

The rate is indicated by slope of the tangent to a curve at time t.This slope is considered as derivative of \(\left( {NO} \right)\)with time.So the rate can be calculated in the following way .

\(Rate{\rm{ }}of{\rm{ }}the{\rm{ }}reaction{\rm{ }}which{\rm{ }}is{\rm{ }}instantaneous\;\; = \;\dfrac{{{{\left( {NO} \right)}_0} - {{\left( {NO} \right)}_t}}}{{\Delta t}}\)

Rate at the time 200 seaconds can be obtained by drawing tangent to the curve at time 200 seaconds.

Yellow line shows the tangent of the graph

\(\begin{aligned}{}Rate\;of\;the\;reaction\; = \;\dfrac{{0.344 - 0.291}}{{250 - 150}}\\ = \;\dfrac{{0.0053}}{{100}}\\ = \;5.3\; \times {10^{ - 5}}\;mol\;{L^{ - 1}}\end{aligned}\)

Hence instantaneous rate of reaction at 100 seacond is \(5.3\; \times {10^{ - 5}}\;mol\;{L^{ - 1}}\).