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JEE Advanced 2023 Paper 1 Online
Numerical
+4
-0
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The plot of $\log k_f$ versus $1 / T$ for a reversible reaction $\mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{P}(\mathrm{g})$ is shown.

JEE Advanced 2023 Paper 1 Online Chemistry - Equilibrium Question 1 English
Pre-exponential factors for the forward and backward reactions are $10^{15} \mathrm{~s}^{-1}$ and $10^{11} \mathrm{~s}^{-1}$, respectively. If the value of $\log K$ for the reaction at $500 \mathrm{~K}$ is 6 , the value of $\left|\log k_b\right|$ at $250 \mathrm{~K}$ is ______.

$$ \begin{aligned} & {[K=\text { equilibrium constant of the reaction }} \\\\ & k_f=\text { rate constant of forward reaction } \\\\ & \left.k_b=\text { rate constant of backward reaction }\right] \end{aligned} $$
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2
JEE Advanced 2022 Paper 2 Online
Numerical
+3
-1
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Concentration of $\mathrm{H}_{2} \mathrm{SO}_{4}$ and $\mathrm{Na}_{2} \mathrm{SO}_{4}$ in a solution is $1 \mathrm{M}$ and $1.8 \times 10^{-2} \mathrm{M}$, respectively. Molar solubility of $\mathrm{PbSO}_{4}$ in the same solution is $\mathrm{X} \times 10^{-\mathrm{Y}} \mathrm{M}$ (expressed in scientific notation). The value of $Y$ is ________.

[Given: Solubility product of $\mathrm{PbSO}_{4}\left(K_{s p}\right)=1.6 \times 10^{-8}$. For $\mathrm{H}_{2} \mathrm{SO}_{4}, K_{a l}$ is very large and $\left.K_{a 2}=1.2 \times 10^{-2}\right]$

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3
JEE Advanced 2022 Paper 1 Online
Numerical
+3
-0
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A solution is prepared by mixing $0.01 \mathrm{~mol}$ each of $\mathrm{H}_{2} \mathrm{CO}_{3}, \mathrm{NaHCO}_{3}, \mathrm{Na}_{2} \mathrm{CO}_{3}$, and $\mathrm{NaOH}$ in $100 \mathrm{~mL}$ of water. $p \mathrm{H}$ of the resulting solution is _________.

[Given: $p \mathrm{~K}_{\mathrm{a} 1}$ and $p \mathrm{~K}_{\mathrm{a} 2}$ of $\mathrm{H}_{2} \mathrm{CO}_{3}$ are $6.37$ and 10.32, respectively; $\log 2=0.30$ ]

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4
JEE Advanced 2020 Paper 2 Offline
Numerical
+4
-0
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A solution of 0.1 M weak base (B) is titrated with 0.1 M of a strong acid (HA). The variation of pH of the solution with the volume of HA added is shown in the figure below. What is the pKb of the base? The neutralisation reaction is given by

$$B + HA\buildrel {} \over \longrightarrow B{H^ + } + {A^ - }$$

JEE Advanced 2020 Paper 2 Offline Chemistry - Equilibrium Question 9 English
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