Question

For a reaction represented by $A\to B$, which of the following options give the value of order of the reaction.
$ROR=$ Rate of reaction
$C=$ Concentration of reactant $A$
$C_0=$ Concentration of reactant $A$ at $t=0$
$P=$ Concentration of product $B$
$t_{\frac{1}{2}}=$ Half life of $A$

• A. $\frac{\log (ROR{)}_2-\log (ROR{)}_1}{\log C_2-\log C_1}$
• B. $1+\left[\frac{\log t_{1/2}^{\prime }-\log t_{1/2}}{\log C_0-\log C_0^{\prime }}\right]$
• C. Slope of $\ln ROR$ vs $\ln C$ graph
• D. "$y-$ intercept" of $\ln ROR$ vs $\ln C$ graph

Answer 1

Answer: (A), (B), (C)

Slope of $\ln ROR$ vs $\ln C$ graph

a)
$ROR=K{C}^n$
$\Rightarrow \frac{(ROR{)}_2}{(ROR{)}_1}={\left(\frac{C_2}{C_1}\right)}^n$
$\Rightarrow n=\frac{\ln (ROR{)}_2-\ln (ROR{)}_1}{\ln C_2-\ln C_1}$

b)
We know,
$t_{1/2}\propto \frac{1}{C_0^{n-1}}$
$\Rightarrow \frac{(t_{1/2}{)}^{\prime }}{\left(t_{1/2}\right)}=\frac{(C_0{)}^{n-1}}{(C_0^{\prime }{)}^{n-1}}$
$\Rightarrow \log (t_{1/2}{)}^{\prime }-\log (t_{1/2})=(n-1)[\log C_0-\log C_0^{\prime }]$
$\Rightarrow n=\frac{\log (t_{1/2}{)}^{\prime }-\log (t_{1/2})}{\log C_0-\log C_0^{\prime }}+1$

c)
$ROR=K{C}^n$
$\Rightarrow \ln \left(ROR\right)=\ln K+n\ln C$
Slope of $\ln \left(ROR\right)$ v/s $\ln \left(C\right)=n$

d)
$ROR=K{C}^n$
$\Rightarrow \ln \left(ROR\right)=\ln K+n\ln C$
$y-$ intercept of $\ln \left(ROR\right)$ v/s $\ln \left(C\right)$
graph is $\ln K$