Ammonium carbamate (NH_4COONH_2) when heated | vedclass

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(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

$0.25$

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(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?

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