The slope of the plot of log p vs 1Tfor solid CO2, is -1538 K. The enthalpy of sublimation of solid CO2, (in KJ mol-1) is

−ΔH2.303R=−1538or ΔH=1538×2.303×8.314×10−3=29.45kJmol−1

The slope of the plot of log p vs 1Tfor solid CO2, is -1538 K. The enthalpy of sublimation of solid CO2, (in KJ mol-1) is

−ΔH2.303R=−1538or ΔH=1538×2.303×8.314×10−3=29.45kJmol−1

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The slope of the plot of log p vs $\frac{1}{T}$ for solid CO₂, is -1538 K. The enthalpy of sublimation of solid CO₂, (in KJ mol^-1) is

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29.45

32.36

35.84

25.68

answer is D.

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$- \frac{ΔH}{2 .303 R} = - 1538$
or $ΔH = 1538 \times 2 .303 \times 8 .314 \times 10^{- 3} = 29.45 {kJmol}^{- 1}$

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The slope of the plot of log p vs 1Tfor solid CO2, is -1538 K. The enthalpy of sublimation of solid CO2, (in KJ mol-1) is

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The slope of the plot of log p vs $\frac{1}{T}$ for solid CO₂, is -1538 K. The enthalpy of sublimation of solid CO₂, (in KJ mol^-1) is