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Find out the value of (P) in the following cycle

Given $Δ H_{1} = - 400 kJ, I . E = + 50 kJ, D = + 50 kJ,$ $Δ H_{e g} = - 30 kJ,$ lattice energy $= - 100 kJ$

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detailed solution

Correct option is C

The following cycle is obtained,

$Δ H_{1} = D + IE + Δ H_{eg} + P + Δ H_{LE}$

$P = ∆ H_{1} - D - I E - ∆ H e q - ∆ H_{L E}$

$P = - 400 - 50 - 50 + 30 + 100 - 500 + 130$

$P = - 370$

Hence, option C is correct.

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Find out the value of (P) in the following cycle Given ΔH1=−400kJ,I.E=+50kJ,D=+50kJ, ΔHeg=−30kJ, lattice energy =−100kJ