Answer to Question #231001 in Chemical Engineering for Lucky

Question

a) A gas A, decomposes ineversibly to form a gas. C, as per the reactionA→2C. The decomposition of A is a first order reaction and is carried out in an isothermal constant pressure and volume batch reactors separately. Derive the expression for the pressure and volume (both separately) of the system as a function of time, assuming gas behave ideally.

Answer 1

This is the first order irreversible gas-phase reaction

$-r_A=\frac{-1}{v}\frac{dN_A}{dt}\\ \therefore N_A= N_A(1-X_A)$

For constant pressure isothermal reaction

$v=v_0(1+ \epsilon_A *A)\\ -r_A=\frac{1}{v_0(1+ \epsilon_A *A)} \frac{dN_A}{dt}(1-x_A)\\ -r_A=\frac{N_Ao}{v_0(1+ \epsilon_A *A)} \frac{dx_A}{dt}\\ -r_A=\frac{C_Ao}{(1+ \epsilon_A *A)} \frac{dx_A}{dt}\\ \implies \frac{C_Ao}{(1+ \epsilon_A *x_A)} \frac{dx_A}{dt}=kC_A\\ \frac{dx_A}{dt}= k \frac{C_Ao(1-x_A)}{(1+ \epsilon_A *x_A)}\\ \frac{dx_A}{(1-x_A)}= k dt\\ t= 0; X_A=0\\ t= t; X_A=X_A\\ -\ln(1-X_A)= kt +C \space but C=0\\ \ln(1-X_A)= kt \\ V-V_o= V_0 \epsilon_AX_A\\ X_A= \frac{V-V_0}{V_0 \epsilon_A}\\ \implies \ln(1-\frac{V-V_0}{V_0 \epsilon})= kt$