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### For the reaction $2NO_2 (g) \rightleftharpoons N_2O_4 (g)$ ....

For the reaction $2NO_2 (g) \rightleftharpoons N_2O_4 (g)$, when $\Delta S = -176.0 JK^{-1}$ and $\Delta H = -57.8 kJ mol^{-1}$, the magnitude of $\Delta G$ at 298 K for the reaction is _ _ _ _ $kJ mol^{-1}$. (Nearest integer)

Solution

We have, $\Delta G = \Delta H - T \Delta S$

$\Delta S = -176.0 JK^{-1} = -0.176 kJ K^{-1}$

$\Delta G = -57.8 - 298 \times (-0.176) \approx -57.8+52.5 = -5.3$

$\therefore |\Delta G | = 5.3 kJ mol^{-1}$

Ans: 5

### A man starts walking from the point P (-3, 4) ....

A man starts walking from the point P (-3, 4), touches the x-axis at R, and then turns to reach at the point Q (0, 2). The man is walking at a constant speed. If the man reaches the point Q in the minimum time, then $50 [(PR)^2 + (RQ)^2 ]$ is equal to _ _ _ _ . Solution For time to be minimum at constant speed, the directions must be symmetric. In other words, the angles made by PR and RQ with the vertical must be the same just like in the law of reflection in optics. $tan \theta = \frac {MP}{MR} = \frac {NQ}{NR}$ $\Rightarrow \frac {3-r}{4} = \frac {r}{2}$ $\Rightarrow r=1$ So, $R \equiv ( - 1,0)$ Now, $50(PR^2+RQ^2)=50[(4+16)+(1+4)]=1250$