Question

The gaseous reaction $A_{\left(g\right)}\to 2B_{\left(g\right)}+C_{\left(g\right)}$ is found to be first order with respect to $A$. If the reaction is started with $p_A=90mmHg$, the pressure after $10$ minutes found to be $180mmHg$. The rate constant of the reaction is?

• A. $1.15\times {10}^{-3}{s}^{-1}$
• B. $2.30\times {10}^{-3}{s}^{-1}$
• C. $3.45\times {10}^{-3}{s}^{-1}$
• D. $4.60\times {10}^{-3}{s}^{-1}$

Answer 1

Answer: (A)

$1.15\times {10}^{-3}{s}^{-1}$

The gaseous reaction,

$\underset{9090-x}{A\left(g\right)}\to \underset{02x}{2B\left(g\right)}+\underset{0x}{C\left(g\right)}$ obeys first order kinetics.

so, total pressure after 10 minutes , $P_T=90-x+2x+x=90+2x$

but given, pressure after 10 minutes = 180mm

so, 90 + 2x = 180

2x = 90

x = 45mm

so, Final pressure , $P_A$ = 90- x = 90mm - 45mm = 45mm

now applying first order reaction formula,

$k=\frac{2.303}{t}log\left(\frac{P_i}{P_f}\right)$

$=\frac{2.303}{10\times 60}log\left(\frac{90}{45}\right)$

$=\frac{2.303}{600}\times log2$

$=1.15\times {10}^{-3}se{c}^{-1}$

hence, rate constant of reaction is $1.15\times {10}^{-3}se{c}^{-1}$