Question

For a zero order chemical reaction,
$2N{H}_3\left(g\right)\to N_2\left(g\right)+3H_2\left(g\right)$
rate of reaction = 0.1 atm /sec. Initially only $N{H}_3$ is present and its pressure = 3atm. Calculate total pressure at t = 10 sec.

Answer 1

$2N{H}_3\left(g\right)⟶N_2\left(g\right)+3H_2\left(g\right)$

$t=0$, $3atm$ $0$ $0$ rate= $0.1atm/sec$

$t=10$, $3-2x$ $x$ $3x$

Given order= zero

$\therefore$ Rate= $k$= rate constant= $0.1$

$\Rightarrow k=\frac{\left[R_o\right]-\left[R\right]}{t}$

$\Rightarrow 0.1=\frac{3-\left[R\right]}{10}\Rightarrow \left[R\right]=2atm⟶\left(1\right)$

Total pressure= $3-2x+x+3x$

$=3+2x⟶\left(2\right)$

At $t=10,\left[R\right]=3-2x=2$ (from (1)) $\Rightarrow x=0.5$

Substitute $x$ in equation $\left(2\right)\Rightarrow$ Total pressure= $3+2\left(0.5\right)=4atm$ at $t=10$