JEE MAIN - Chemistry (2021 - 31st August Morning Shift - No. 14)
Consider the following cell reaction :
$$C{d_{(s)}} + H{g_2}S{O_{4(s)}} + {9 \over 5}{H_2}{O_{(l)}}$$ $$\rightleftharpoons$$ $$CdS{O_4}.{9 \over 5}{H_2}{O_{(s)}} + 2H{g_{(l)}}$$
The value of $$E_{cell}^0$$ is 4.315 V at 25$$^\circ$$C. If $$\Delta$$H$$^\circ$$ = $$-$$825.2 kJ mol$$-$$1, the standard entropy change $$\Delta$$S$$^\circ$$ in J K$$-$$1 is ___________. (Nearest integer) [Given : Faraday constant = 96487 C mol$$-$$1]
$$C{d_{(s)}} + H{g_2}S{O_{4(s)}} + {9 \over 5}{H_2}{O_{(l)}}$$ $$\rightleftharpoons$$ $$CdS{O_4}.{9 \over 5}{H_2}{O_{(s)}} + 2H{g_{(l)}}$$
The value of $$E_{cell}^0$$ is 4.315 V at 25$$^\circ$$C. If $$\Delta$$H$$^\circ$$ = $$-$$825.2 kJ mol$$-$$1, the standard entropy change $$\Delta$$S$$^\circ$$ in J K$$-$$1 is ___________. (Nearest integer) [Given : Faraday constant = 96487 C mol$$-$$1]
Answer
25
Explanation
$$\Delta G^\circ = - nFE^\circ = \Delta H^\circ - T\Delta S^\circ $$
$$ \therefore $$ $$\Delta$$S$$^\circ$$ $$ = {{\Delta H^\circ + nFE^\circ } \over T}$$
$$ = {{( - 825.2 \times {{10}^3}) + (2 \times 96487 \times 4.315)} \over {298}}$$
$$ = {{ - 825.2 \times {{10}^3} + 832.682 \times {{10}^3}} \over {298}}$$
$$ = {{7.483 \times {{10}^3}} \over {298}} = 25.11$$ JK$$-$$1 mol$$-$$1
$$\therefore$$ Nearest integer answer is 25.
$$ \therefore $$ $$\Delta$$S$$^\circ$$ $$ = {{\Delta H^\circ + nFE^\circ } \over T}$$
$$ = {{( - 825.2 \times {{10}^3}) + (2 \times 96487 \times 4.315)} \over {298}}$$
$$ = {{ - 825.2 \times {{10}^3} + 832.682 \times {{10}^3}} \over {298}}$$
$$ = {{7.483 \times {{10}^3}} \over {298}} = 25.11$$ JK$$-$$1 mol$$-$$1
$$\therefore$$ Nearest integer answer is 25.
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