Chapter 6, Problem 6A.1ST

Interpretation Introduction

Interpretation:

The relation between the rate of formation of $N{H}_4^{+}$ and the rate of consumption of ${(N{H}_2)}_{2}CO$ has to be derived.  

Concept Introduction:

Rate of the reaction is the change in the concentration of reactant or a product with time. It can be varied in accordance with temperature, pressure, concentration, presence of catalyst, surface area…

For a general reaction,

$aA+bB\to cC+dD$

$\text{Rate}\text{=}\text{-}\frac{\text{1}}{\text{a}}\frac{\text{Δ}\left[\text{A}\right]}{\text{Δt}}\text{=}\text{-}\frac{\text{1}}{\text{b}}\frac{\text{Δ}\left[\text{B}\right]}{\text{Δt}}\text{=}\frac{\text{1}}{\text{c}}\frac{\text{Δ}\left[\text{C}\right]}{\text{Δt}}\text{=}\frac{\text{1}}{\text{d}}\frac{\text{Δ}\left[\text{D}\right]}{\text{Δt}}$

The negative sign indicates the reduction of concentration of reactant.

$\Delta \left[A\right],\Delta \left[B\right],\Delta \left[C\right]and\Delta \left[D\right]$ are the change in concentration of A,B, C and D over time period $\Delta t$

Explanation of Solution

The balanced equation is,

${(N{H}_4)}_{2}CO+H_{2}O(l)\stackrel{H^{+}}{\to }2N{H}_4^{+}(aq)+C{O}_3^{2-}(aq)$

Rate consumption of reactants = Rate of formation of product.

$\frac{1}{2}\left(\frac{\Delta [N{H}_4^{+}]}{\Delta t}\right)=-\left(\frac{\Delta {(N{H}_2)}_{2}CO}{\Delta t}\right)$

Hence,

$\left(\frac{\Delta [N{H}_4^{+}]}{\Delta t}\right)=-2\times \left(\frac{\Delta {(N{H}_2)}_{2}CO}{\Delta t}\right)$

Therfore, the positive rate formation of $N{H}_4^{+}=2\times rateconsumptionof{(N{H}_4)}_{2}CO$