JEE MAIN - Chemistry (2019 - 12th April Morning Slot - No. 17)
Enthalpy of sublimation of iodine is 24 cal g–1
at 200 oC. If specific heat of I2(s) and l2 (vap) are 0.055 and
0.031 cal g–1K
–1
respectively, then enthalpy of sublimation of iodine at 250 oC in cal g–1
is :
2.85
22.8
11.4
5.7
Explanation
Kirchoff's Equation :
$${{\Delta {H_{{T_2}}} - \Delta {H_{{T_1}}}} \over {{T_2} - {T_1}}} = \Delta \left( {{C_p}} \right)$$
Sublimation of iodine :
I2(s) $$ \to $$ I2(g)
$${{\Delta {H_{250^\circ C}} - \Delta {H_{200^\circ C}}} \over {250 - 200}} = {C_p}\left[ {{I_2}(g)} \right] - {C_p}\left[ {{I_2}(s)} \right]$$
$${{\Delta {H_{250^\circ C}} - 24} \over {50}} = \left( {0.031 - 0.055} \right)$$
$$ \Rightarrow $$ $${\Delta {H_{250^\circ C}}}$$ = 22.8 cal g–1
$${{\Delta {H_{{T_2}}} - \Delta {H_{{T_1}}}} \over {{T_2} - {T_1}}} = \Delta \left( {{C_p}} \right)$$
Sublimation of iodine :
I2(s) $$ \to $$ I2(g)
$${{\Delta {H_{250^\circ C}} - \Delta {H_{200^\circ C}}} \over {250 - 200}} = {C_p}\left[ {{I_2}(g)} \right] - {C_p}\left[ {{I_2}(s)} \right]$$
$${{\Delta {H_{250^\circ C}} - 24} \over {50}} = \left( {0.031 - 0.055} \right)$$
$$ \Rightarrow $$ $${\Delta {H_{250^\circ C}}}$$ = 22.8 cal g–1
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