JEE Advanced Previous Year Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

*Simulator*

**Previous Years JEE Advance Questions**

below. The plot that follows Arrhenius equation is –

**[JEE 2010]**

**Sol.**(A)

Arrhenius equation : $\mathrm{k}=\mathrm{Ae}^{-\mathrm{E}_{\mathrm{a}} / \mathrm{RT}}$

(taking A and $\mathrm{E}_{\mathrm{a}}$ to be constant, differentiating w.r.t. T)

$\frac{\mathrm{d} \mathrm{k}}{\mathrm{d} \mathrm{T}}=\left[\frac{\mathrm{AE}_{\mathrm{a}}}{\mathrm{RT}^{2}}\right] \mathrm{e}^{-\mathrm{E}_{\mathrm{a}} / \mathrm{RT}}$

assuming A and $\mathrm{E}_{\mathrm{a}}$ to be constant theoretically, the plot should be

ButNo such option is given. Now experimentally, A and $\mathrm{E}_{\mathrm{a}}$ both vary with temperature and k increases as T increases and become very large at infinite temperature. Hence option (A) is correct.

**[JEE 2010]**

**Sol.**0

Zero order reaction because rate is constant with time.

Alternative solution : By hit and trial method, assuming the reaction is of zero order, putting given data in integrated expression for zero order.

the value of k is same from different data so reaction is zero order reaction.

$2 \mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) \longrightarrow 4 \mathrm{NO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})$

(A) the concentration of the reactant decreases exponentially with time

(B) the half-life of the reaction decreases with increasing temperature.

(C) the half-life of the reaction depends on the initial concentration of the reactant.

(D) the reaction proceeds to 99.6% completion in eight half-life duration.

**[JEE 2011]**

**Sol.**(A,B,D)

$\mathrm{A}_{\mathrm{t}}=\mathrm{A}_{0} \mathrm{e}^{-\mathrm{kt}}$

for option (D)

$\frac{1}{\mathrm{t}_{1 / 2}} \ln \frac{100}{50}=\frac{1}{\mathrm{t}_{99.6 \%}} \ln \frac{100}{0.4}$

\mathrm{t}_{1 / 8} \text { and } \mathrm{t}_{1 / 10}

$$ What is the value of $$

\frac{\left[\mathrm{t}_{1 / 8}\right]}{\mathrm{t}_{1 / 10}} \times 10 ?\left(\text { take } \log _{10} 2=0.3\right)

$$

**[JEE 2012]**

**Sol.**9

$$

\mathrm{P}+\mathrm{Q} \longrightarrow \mathrm{R}+\mathrm{S}

$$

the time taken for 75% reaction of P is twice the time taken for 50% reaction of P. The

concentration of Q varies with reaction time as shown in the figure. The overall order of the

reaction is –

(A) 2 (B) 3 (C) 0 (D) 1

**[JEE 2013]**

**Sol.**9

\mathrm{M} \rightarrow \mathrm{N}$$, the rate of disappearance of M increases by a factor of 8 upon doubling the concentration of M. The order of the reaction with respect to M is

(A) 4 (B) 3(C) 2(D) 1

**[JEE 2014]**

**Sol.**(D)

\mathrm{H}_{2} \mathrm{SO}_{4}$$ , the complex diaquodioxalatoferrate(II) is oxidized by $\mathrm{MnO}_{4}^{-}$.For this reaction, the ratio of the rate of change of $$

\left[\mathrm{H}^{+}\right]$$ to the rate of change of [MnO4–] is.

**[JEE 2015]**

**Sol.**(B)

If this reaction is conducted at $\left(P, T_{2}\right)$, with $\mathrm{T}_{2}>\mathrm{T}_{1}$, the % yield of ammonia as a function of time is represented by –

**[JEE 2015]**

**Sol.**(B)

$\therefore$ % yield will increase in initial stages due to increase in net speed

As time proceeds $\Rightarrow \mathrm{r}_{\mathrm{net}}=\mathrm{k}_{\mathrm{f}}\left[\mathrm{N}_{2}\right]\left[\mathrm{H}_{2}\right]^{3}-\mathrm{k}_{\mathrm{b}}\left[\mathrm{NH}_{3}\right]^{2}$

On increasing temp., $\mathrm{k}_{\mathrm{f}} \& \mathrm{k}_{\mathrm{b}}$ increase

but increase of $\mathrm{k}_{\mathrm{b}}$ is more

so % yield will decrease

% yield will increase in initial stage due to enhance speed but as time proceeds , final yield is governed by thermodynamics due to which yield decrease since reaction is exothermic

total pressure at the beginning (t = 0) and at time t are P0 and Pt, respectively. Initially, only A

is present with concentration [A]0, and $t_{1 / 3}$ is the time required for the partial pressure

of A to reach $1 / 3^{\mathrm{rd}}$ of its initial value. The correct option(s) is (are) :-

(Assume that all these gases behave as ideal gases)

**[JEE Advance 2018]**

**Sol.**(A,D)

$\mathrm{A}(\mathrm{g})+\mathrm{B}(\mathrm{g}) \square \mathrm{AB}(\mathrm{g})$

The activition energy of the backward reaction exceeds that of the forward reaction by 2RT $\left(\text { in } \mathrm{J} \mathrm{mol}^{-1}\right)$. If the pre-exponential factor of the forward reaction is 4 times that of the reverse reaction, the absolute value of $\Delta \mathrm{G}^{\theta}$(in J mol–1) for the reaction at 300 K is____.

(Given ; ln (2) = 0.7, RT = 2500 J $\mathrm{mol}^{-1}$ at 300 K and G is the Gibbs energy)

**[JEE Advance 2018]**

**Sol.**8500