91Ó°ÊÓ

\(I = \frac{Q}{t}\)

• Here, I is the current, Q is the charge, and t is the time.
• The SI unit of current is Ampere, denoted by symbol A.
• Thus, 1A current means 1C of charge passes through a wire every 1 second. Or,\({\bf{1\;A = }}\frac{{{\bf{1C}}}}{{{\bf{1\;s}}}}\)

The current passing through the wire is 2.5A and time is 35 min.

Calculation:

The current is related to charge and time as follows:

\({\rm{I = }}\frac{{\rm{Q}}}{{\rm{t}}}\)

Thus, charge can be calculated from current as:

\({\rm{Q = I \times t}}\)

Putting the values,

\(\begin{array}{l}{\rm{Q = (2}}{\rm{.5\;A)(35\;min)}}\left( {\frac{{{\rm{60\;s}}}}{{{\rm{1\;min}}}}} \right){\rm{ = 5250A}}\\{\rm{s = 5250C}}\end{array}\)

\(\Delta G_{f\left( {N{O_L}} \right)}^^\circ = 51.30\;{\rm{kJ}}/{\rm{mol}}\)

\(\Delta {G_{f\left( {\;{{\rm{N}}_2}{{\rm{O}}_4}} \right)}} = 99.8\;{\rm{kJ}}/{\rm{mol}}\)

We need to calculate the equilibrium constant for the dissociation of dintrogen tetroxide \(\left( {{{\rm{N}}_2}{{\rm{O}}_4}} \right)\)

\(\Delta {G^^\circ } = \sum \Delta G_f^^\circ \)(products) \( - \sum \Delta G_f^^\circ \) (reactants)

\( = 2\Delta G_{{N_2}}^^\circ - \Delta {G^^\circ }^^\circ {O_4}\)

\( = 2 \times 51.3 - 99.8\)

\( = 2.8\;{\rm{kJ}}/{\mathop{\rm mo}\nolimits} l\quad (1\;{\rm{kJ}} = 1000\;{\rm{J}})\)

or\(2800\;{\rm{J}}/\) mol

\(\Delta {G^^\circ } = - RT\ln {K_{eq}}\)

\( = \frac{{ - 2800}}{{8.314 \times 298}}\)

\(\ln {k_{{\rm{eq }}}} = - 1.130\)

or\({K_{eq}} = {e^{ - 1.130}}\)

\({K_{{\rm{eq }}}} = 0.30\)