Enthalpy of sublimation of iodine is 24 cal g-1 at 200°C. If specific heat of I2(s) and I2(vap) are 0.055 and 0.031 cal g-1K-1 respectively, then enthalpy of sublimation of iodine at 250°C in cal g-1 is :

Question

Enthalpy of sublimation of iodine is 24 cal g-1 at 200°C. If specific heat of I2(s) and I2(vap) are 0.055 and 0.031 cal g-1K-1 respectively, then enthalpy of sublimation of iodine at 250°C in cal g-1 is : (A) 2.85 (B) 11.4 (C) 5.7 (D) 22.8

Tardigrade Admin · Accepted Answer

I2(s) arrow I2(g): Δ H1 = 24 cal/g at 200° C Δ H2 = Δ H1 + Δ CPrxn (T2-T1) = 24 + (0.031 -0.055) × 50 = 24-1.2 = 22.8 Cal/g

Q. Enthalpy of sublimation of iodine is $24 c a l g^{- 1}$ at $20 0^{\circ} C$ . If specific heat of $I_{2} (s)$ and $I_{2} (v a p)$ are $0.055$ and $0.031 c a l g^{- 1} K^{- 1}$ respectively, then enthalpy of sublimation of iodine at $25 0^{\circ} C$ in $c a l g^{- 1}$ is :

2.85

11.4

5.7

22.8

$I_{2 (s)} \to I_{2 (g)} : Δ H_{1} = 24 c a l / g a t 20 0^{\circ} C$
$Δ H_{2} = Δ H_{1} + Δ C_{P_{r x n}} (T_{2} - T_{1})$
= $24 + (0.031 - 0.055) \times 50$
= $24 - 1.2$
= $22.8 C a l / g$

Q. Enthalpy of sublimation of iodine is 24calg−1 at 200∘C. If specific heat of I2​(s) and I2​(vap) are 0.055 and 0.031calg−1K−1 respectively, then enthalpy of sublimation of iodine at 250∘C in calg−1 is :

2.85

11.4

5.7

22.8

I2(s)​→I2(g)​:ΔH1​=24cal/gat200∘C ΔH2​=ΔH1​+ΔCPrxn​​(T2​−T1​) = 24+(0.031−0.055)×50 = 24−1.2 = 22.8Cal/g

Q. Enthalpy of sublimation of iodine is $24 c a l g^{- 1}$ at $20 0^{\circ} C$ . If specific heat of $I_{2} (s)$ and $I_{2} (v a p)$ are $0.055$ and $0.031 c a l g^{- 1} K^{- 1}$ respectively, then enthalpy of sublimation of iodine at $25 0^{\circ} C$ in $c a l g^{- 1}$ is :

$I_{2 (s)} \to I_{2 (g)} : Δ H_{1} = 24 c a l / g a t 20 0^{\circ} C$
$Δ H_{2} = Δ H_{1} + Δ C_{P_{r x n}} (T_{2} - T_{1})$
= $24 + (0.031 - 0.055) \times 50$
= $24 - 1.2$
= $22.8 C a l / g$