Ammonium carbamate (NH_4COONH_2) when heated | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The dissociation reaction is: $NH_4COONH_2(s) \rightleftharpoons 2NH_3(g) + CO_2(g)$.The number of moles of gaseous products produced per mole of

Vedclass · Accepted Answer

$0.25$

Vedclass · Answer

(B) The dissociation reaction is: $NH_4COONH_2(s) ightleftharpoons 2NH_3(g) + CO_2(g)$.The number of moles of gaseous products produced per mole of reactant dissociated is $n = 2 + 1 = 3$.The theoretical vapour density $(D_t)$ is calculated as: $D_t = \frac{	ext{Molar Mass}}{2} = \frac{78}{2} = 39$.Given the observed vapour density at equilibrium is $D_0 = 26$.The degree of dissociation $(\alpha)$ is given by the formula: $\alpha = \frac{D_t - D_0}{D_0(n - 1)}$.Substituting the values: $\alpha = \frac{39 - 26}{26(3 - 1)} = \frac{13}{26 	imes 2} = \frac{13}{52} = 0.25$.

Ammonium carbamate $(NH_4COONH_2)$ when heated to $200 \ ^oC$ gives a mixture of $NH_3$ and $CO_2$ with a vapour density of $26$ at equilibrium. What is the degree of dissociation of ammonium carbamate?

Similar Questions

At a certain temperature,the vapour density of $N_2O_4$ is $30$. Find the percentage of dissociation of $N_2O_4$ at this temperature.

In a $Victor \ Meyer$ experiment,$0.23 \ g$ of a volatile substance displaces $112 \ mL$ of air at $STP$. Calculate the vapor density of the substance.

If the vapor density of a volatile substance relative to $CH_4$ is $4$,then find the molecular mass of that substance.

If the density of air at $STP$ is $0.001293 \, g \, mL^{-1}$,what will be its vapor density?

The density of air is $0.00130 \ g/mL$. The vapour density of air will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module