Question

Some solid $NH_{4}HS$ is placed in a flask containing 0.5 moles of $NH_{3}$. What would be the pressure of $NH_{3}$(g) and $H_{2}S$(g) when equilibrium is reached? $NH_{4}HS(s) \rightleftharpoons NH_{3}(g)+H_{2}S(g); K_{p} = 0.11$.

Answer 1

HEY MATE YOUR ANSWER IS

Aromatic amine( amines with benzene group) are always less basic then normal aliphatic amines and even ammonia(NH₃).

Due to +inductive effect from alkyl groups, amines are more basic than Ammonia.....that is 
                                 CH₃NH₂ > NH₃ in basic strength.

while aromatic amines are least basic(eg: C₆H₅NH₂)

Therefore the order of basic strength is ,

                             CH₃NH₂ > NH₃ > C₆H₅NH₂


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Answer 2

"The given compound is ${ NH }_{ 4 }HS$

The decomposition reaction is as follows.

$\begin{matrix} { NH }_{ 4 }HS \\ Initial \\ Equilibri \end{matrix}\quad \rightarrow \quad \begin{matrix} { NH }_{ 3(g) } \\ 0.5\quad atm \\ 0.5+x\quad atm \end{matrix}\quad +\quad \begin{matrix} { H }_{ 2 }{ S }_{ (g) } \\ 0\quad atm \\ x\quad atm \end{matrix}$

Then,

Let's find out the "x" value

$0.5\quad +\quad x\quad +\quad x\quad =\quad 2x\quad +\quad 0.5\quad =\quad 0.84(given)$

$x\quad =\quad 0.17\quad atm\\ { P }_{ { NH }_{ 3 } }\quad =\quad 0.5\quad +0.17\quad =\quad 0.67\quad atm$

${ P }_{ { H }_{ 2 }S }\quad =\quad 0.17\quad atm$

$K\quad =\quad { P }_{ { NH }_{ 3 } }\quad \times \quad { P }_{ { H }_{ 2 }S }$

$=\quad 0.67\quad \times \quad 0.17\quad at{ m }^{ 2 }$

$K\quad =\quad 0.1139$"