Question

$A_{(g)}\stackrel{\Delta }{\to }{P}_{(g)}+Q_{(g)}+R_{(g)}$, follows first order kinetics with a half life of $69.3s$ at $500^{\circ }C$. Starting from the gas ${}^{\prime }{A}^{\prime }$ enclosed in a container at $500^{\circ }C$ and at a pressure of $0.4atm$, the total pressure of the system after $230s$ will be

• A. 1.32 atm
• B. 1.12 atm
• C. 1.15 atm
• D. 1.22 atm

Answer 1

Answer: (B)

$p_t=(0.4-x)+x+x+x$

$=0.4+2x$

or $x=\frac{p_t-0.4}{2}$

$\therefore p_A=p_0-x=p_0-\frac{p_t-0.4}{2}$

$=\frac{2\times 0.4-p_t+0.4}{2}$

$=\frac{1.2-p_t}{2}$

From $k=\frac{2.303}{t}\log \left(\frac{p_0}{p_A}\right)$

$\frac{0.693}{t_{1/2}}=\frac{2.303}{230}\log \left(\frac{0.4\times 2}{1.2-p_t}\right)$

$\frac{0.693}{69.3}=\frac{2.303}{230}\log \left(\frac{0.8}{1.2-p_t}\right)$

$\log \left(\frac{0.8}{1.2-p_t}\right)=\frac{0.693\times 230}{69.3\times 2.303}$

$=0.9987$

$\frac{0.8}{1.2-p_t}=$ antilog (0.9987)

$\frac{0.8}{1.2-p_t}\approx 10$

$\therefore 12-10p_t=0.8$

or $p_t=1.12atm$