AP EAMCET 2020 Chemistry Question Paper with Answer and Solution

(A) The relationship between standard Gibbs energy change and equilibrium constant is given by: $\Delta G^{\circ} = -RT \ln K$ or $\Delta G^{\circ} = -2.303 RT \log K$.
Given: $\Delta G^{\circ} = 13.9 \ kJ \ mol^{-1} = 13900 \ J \ mol^{-1}$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$,$T = 295 \ K$.
Substituting the values: $13900 = -2.303 \times 8.314 \times 295 \times \log K$.
$\log K = \frac{-13900}{2.303 \times 8.314 \times 295} = \frac{-13900}{5650.3} \approx -2.46$.
Wait,the standard Gibbs energy change for the formation of urea from ammonia and carbon dioxide is typically negative. Assuming the value provided is $\Delta G^{\circ} = -13.9 \ kJ \ mol^{-1}$:
$\log K = \frac{13900}{5650.3} \approx 2.46$.
$K = 10^{2.46} \approx 288.4 = 2.88 \times 10^2$.
Hence,option $(A)$ is correct.

(B) For the reaction: $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$
Degree of dissociation $\alpha = 0.5$
Total pressure $P = 1 \ atm$
Temperature $T = 60 + 273 = 333 \ K$
The equilibrium constant $K_p$ is given by the expression: $K_p = \frac{4\alpha^2 P}{1-\alpha^2}$
Substituting the values: $K_p = \frac{4(0.5)^2 \times 1}{1-(0.5)^2} = \frac{1}{0.75} = 1.333$
Standard free energy change $\Delta G^\circ = -RT \ln K_p$
$\Delta G^\circ = -8.314 \times 333 \times \ln(1.333) \approx -8.314 \times 333 \times 0.2874 \approx -796.8 \ J \ mol^{-1}$
Given the options,the closest value is $-763.8 \ J \ mol^{-1}$.

(A) Internal energy $(U)$ is a state function.
State functions depend only on the initial and final states of the system and are independent of the path taken to reach the final state.
In both processes $I$ and $II$,the system starts at state $A$ and ends at state $B$.
Therefore,the change in internal energy for both processes is the same: $\Delta U_1 = U_B - U_A$ and $\Delta U_2 = U_B - U_A$.
Hence,$\Delta U_1 = \Delta U_2$.

(B) reaction is spontaneous when the change in Gibbs free energy $( \Delta G )$ is negative.
For a spontaneous process,$ \Delta G < 0 $ at constant temperature and pressure.
If $ \Delta G = 0 $,the reaction is at equilibrium.
If $ \Delta G > 0 $,the reaction is non-spontaneous.

(A) The relationship between standard Gibbs free energy change $\Delta G^0$ and equilibrium constant $K_p$ is given by the formula: $\Delta G^0 = -RT \ln K_p$.
Given: $R = 8.314 \ J \ K^{-1} \ mol^{-1}$,$T = 300 \ K$,$K_p = 1.8 \times 10^{-7}$.
Substituting the values: $\Delta G^0 = -(8.314 \ J \ K^{-1} \ mol^{-1}) \times (300 \ K) \times \ln(1.8 \times 10^{-7})$.
$\Delta G^0 = -2494.2 \times (\ln(1.8) + \ln(10^{-7}))$.
$\Delta G^0 = -2494.2 \times (0.5878 - 16.118)$.
$\Delta G^0 = -2494.2 \times (-15.5302) \approx 38735 \ J \ mol^{-1}$.
Converting to $kJ \ mol^{-1}$: $\Delta G^0 \approx 38.74 \ kJ \ mol^{-1}$.
Thus,the closest option is $38.72 \ kJ \ mol^{-1}$.

(A) The fundamental frequency of a closed organ pipe is $f_c = \frac{v}{4l}$.
The fundamental frequency of an open organ pipe is $f_o = \frac{v}{2l}$.
Given the initial beat frequency is $2 \text{ Hz}$:
$f_o - f_c = 2$
$\frac{v}{2l} - \frac{v}{4l} = 2$
$\frac{v}{4l} = 2 \implies \frac{v}{l} = 8$.
After changing the lengths,the new length of the open pipe is $l' = \frac{l}{2}$ and the new length of the closed pipe is $l'' = 2l$.
The new fundamental frequencies are:
$f_o' = \frac{v}{2l'} = \frac{v}{2(l/2)} = \frac{v}{l} = 8 \text{ Hz}$.
$f_c' = \frac{v}{4l''} = \frac{v}{4(2l)} = \frac{v}{8l} = \frac{8}{8} = 1 \text{ Hz}$.
The new beat frequency is $|f_o' - f_c'| = |8 - 1| = 7 \text{ beats } \sec^{-1}$.

(D) We know that the direction cosines of a line satisfy the relation: $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$,where $\alpha, \beta, \gamma$ are the angles made by the line with the $X, Y,$ and $Z$-axes respectively.
Given $\alpha = \frac{\pi}{3}$ and $\beta = \frac{\pi}{4}$.
Substituting these values into the identity:
$\cos^2 \left(\frac{\pi}{3}\right) + \cos^2 \left(\frac{\pi}{4}\right) + \cos^2 \gamma = 1$
$\Rightarrow \left(\frac{1}{2}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 + \cos^2 \gamma = 1$
$\Rightarrow \frac{1}{4} + \frac{1}{2} + \cos^2 \gamma = 1$
$\Rightarrow \frac{3}{4} + \cos^2 \gamma = 1$
$\Rightarrow \cos^2 \gamma = 1 - \frac{3}{4} = \frac{1}{4}$
$\Rightarrow \cos \gamma = \pm \frac{1}{2}$.
Considering the principal value,$\cos \gamma = \frac{1}{2} = \cos \left(\frac{\pi}{3}\right)$,which gives $\gamma = \frac{\pi}{3}$.