The value of $K_{\mathrm{P}}$ for the process: $\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}(\mathrm{s})\rightleftharpoons \mathrm{CuSO}_4 \cdot 3 \mathrm{H}_2 \mathrm{O}(\mathrm{s})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{g})$ is $1.21 \times 10^{-4} \mathrm{~atm}^2$ at certain temperature. If aqueous tension at that temperature is 40 torr, then at what relative humidity of air will $\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}$ effloresces?

JEE CHEMISTRY MCQ: Equilibrium

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The value of

K_{\mathrm{P}}

for the process:

\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}(\mathrm{s})\rightleftharpoons \mathrm{CuSO}_4 \cdot 3 \mathrm{H}_2 \mathrm{O}(\mathrm{s})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{g})

1.21 \times 10^{-4} \mathrm{~atm}^2

at certain temperature. If aqueous tension at that temperature is 40 torr, then at what relative humidity of air will

\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}

effloresces?

View answer and explanation

Correct option: D

Below

20.9 \%

Option A ([Co(NH3)6]3+) is correct because it forms a low-spin, diamagnetic octahedral complex due to the strong field ligand NH3.

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\mathrm{K}_1

\mathrm{K}_2

\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\quad\ldots

\mathrm{NO}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{~N}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})\quad\ldots

K{a_1},\,\,\,K{a_2}\,\,\,\mathrm{and}\,\,\,K{a_3}

\begin{array}{l} {H_2}S\leftrightharpoons{H^ + } + H{S^ - }\\ H{S^ - }\leftrightharpoons{H^ + } + {S^{2 - }}\\ {H_2}S\leftrightharpoons2{H^ + } + {S^{2 - }} \end{array}

K{a_1},\,\,K{a_2}\,,K{a_3}

\mathrm{pH}

0.1 \mathrm{~M}-\mathrm{N}_2 \mathrm{H}_4

\mathrm{N}_2 \mathrm{H}_4,

K_{{b}_1}=3.6 \times 10^{-6}, K_{{b}_2}=6.4 \times 10^{-12}, \log 2=0.3

\log 3=0.48

{\text{Ba}}{{\text{O}}_2}(s) \rightleftharpoons {\text{BaO}}(s) + {{\text{O}}_2}(g);\Delta H = - {\text{ve}}.

\text{O}_2