Question

For a reaction: \[2{A_{(g)}} \to 3{B_{(g)}} + {C_{(g)}}\] the pressure at different instants is given as shown:

Time(min)	0	13.86	\[\infty \]
Total pressure (mmHg)	50	80	90

Calculate value of rate constant in \[{\min ^{ - 1}}\] ?
[Given ln 2=0.693]

Answer 1

Hint: The rate of a reaction for a given chemical reaction is the measure of the change or decrease in concentration of the reactants or the change or increase in concentration of the products per unit time. The rate of a chemical reaction may be defined mathematically as the change in concentration of a substance divided by the time interval during which this change is observed in the reaction:
\[ \Rightarrow Rate = \dfrac{{\Delta concentration}}{{\Delta time}}\]

Complete answer:
We can say that for this given reaction the pressure at time t is given as

		Pressure
TIME	2A	3B	C
0	\[{P_0}\] = 50	0	0
t	\[{P_0}\] -2x	3x	x

Hence we can say that total pressure is \[{P_0}\]+2x
At time t=13.86 we are given that Total pressure is 80 mmHg
\[ \Rightarrow {P_0} + 2x = 80\]
\[ \Rightarrow 50 + 2x = 80\]
\[ \Rightarrow x = 15\]
We also know that the expression for rate constant, k is written as:
\[ \Rightarrow k = \dfrac{1}{t}\ln \dfrac{{{A_0}}}{{{A_t}}}\]
Where \[{A_0}\] is the initial concentration and \[{A_t}\] is the concentration at time t
\[ \Rightarrow k = \dfrac{1}{{13.86}}\ln \dfrac{{50}}{{50 - 2 \times 15}}\]
\[ \Rightarrow k = \dfrac{1}{{13.86}}\ln \dfrac{{50}}{{20}}\]
\[ \Rightarrow k = 6.6 \times {10^{ - 2}}{\min ^{ - 1}}\]
Therefore we can say that the rate constant of the given reaction is \[6.6 \times {10^{ - 2}}{\min ^{ - 1}}\]

Note:
We use the minus sign before the ratio in the reactant equation because a rate is a positive number. A decrease in reactants is indicated by this minus sign. We do not use the minus sign when calculating the rate of a reaction from the products. The rate law is an expression that expresses the relationship of the rate of a given chemical reaction to the rate constant and the concentrations of reactants raised to some power.