For a gas reaction at T(K) the rate is given by $\frac{{\mathrm{dp}}_{\mathrm{A}}}{\mathrm{dt}}=K^1{P}_A^2$atm/hr.If the rate equation is expressed as: ${\mathrm{r}}_{\mathrm{A}}=-\frac{1}{\mathrm{V}}\frac{{\mathrm{dN}}_{\mathrm{A}}}{\mathrm{dt}}={\mathrm{KC}}_{\mathrm{A}}^2, \mathrm{mol}/(\mathrm{litre}-\mathrm{hr}), \mathrm{the} \mathrm{rate} \mathrm{constant} \mathrm{k} \mathrm{is} \mathrm{given} \mathrm{by} \mathrm{where} \mathrm{R}  \mathrm{ideal} \mathrm{gas} \mathrm{law} \mathrm{constant}, \mathrm{cal}/\mathrm{g} \mathrm{mol} \mathrm{K}$