Home
Class 11
CHEMISTRY
Calculate the equilibrium constant (in m...

Calculate the equilibrium constant (in multiples of `10^(-4)`) for the reaction `PCl_(5(g)) rarr PCl_(3(g)) + Cl_(2(g))` at 400K, if `Delta H^(0) = 77.2KJ "mole"^(-1) and Delta S^(0) = 122JK^(-1) "mole"^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
2

`Delta G^(0) = Delta H^(0) - T Delta S^(0)`
`= 77.2 - (400 xx 122)/(1000) = 28.4 KJ`
`Delta G^(0) = -RT ln K rArr K = 2 xx 10^(-4)`
Promotional Banner

Topper's Solved these Questions

  • CHEMICAL THERMODYNAMICS

    AAKASH SERIES|Exercise PRACTICE SHEET (EXERCISE- I) (LEVEL- I (MAIN) STRAIGHT OBJECTIVE TYPE QUESTIONS)|17 Videos
  • CHEMICAL THERMODYNAMICS

    AAKASH SERIES|Exercise PRACTICE SHEET (EXERCISE- I) (LEVEL- II (ADVANCED) STRAIGHT OBJECTIVE TYPE QUESTIONS)|3 Videos
  • CHEMICAL THERMODYNAMICS

    AAKASH SERIES|Exercise LECTURE SHEET (EXERCISE-III) (LEVEL - II (ADVANCED) MATRIX MATCHING TYPE QUESTIONS)|2 Videos
  • CHEMICAL THERMODYANMICS

    AAKASH SERIES|Exercise Questions For Descriptive Answers|28 Videos
  • ELECTRON MIGRATION EFFECTS

    AAKASH SERIES|Exercise QUESTIONS FOR DESCRIPTIVE ANSWERS|10 Videos

Similar Questions

Explore conceptually related problems

For the reaction PCl_(5(g)) harr PCl_(3(g)) + Cl_(2(g)) . Which of the graph "is"//"are" correct?

Calculate equilibrium constant (in multiples of 10^(-80) ) when sodium reduces Aluminium oxide to aluminium at 298 K. Delta G^(@)_(f) of Na_(2)O_(3(s)) at 298 K = -377 KJ "mole"^(-1) and Delta G^(@)_(f) of Al_(2)O_(3) at 298K = -1582 KJ "mole"^(-1) )

For the reaction PCl_(5(g)) hArr PCl_(3(g)) + Cl_(2(g)) the forward reaction at constant temperature is favoured by

At 27^(@)C, K_(p) value for the reversible reaction PCl_(5)(g)harrPCl_(3)(g)+Cl_(2)(g) is 0.65, calculate K_(c) .

For the reaction X_(2)O_(4(l)) rarr 2XO_(2(g)), Delta U= 2.1 "kcal", Delta S= 20 "cal " K^(-1) at 300K, Hence Delta G is

PCl_(5) (g) leftrightarrow PCl_3(g) +Cl_2(g) At 300K, K_C is 0.204 mol L^-1 . What is the value of K_p ?

At 1000 K, the equilibrium constant, K_(c) for the reaction 2NOCl(g) hArr 2NO(g) +Cl_(2)(g) is 4.0xx10^(-6) mol L^(-1) . The K_(p) (in bar) at the same temperature is (R=0.083" L bar K"^(-1) mol^(-4))

For the reaction at 300 K A_((g)) harr V_((g)) + S_((g) . Delta_(t) H^(@) = - 30 "KJ/mol" Delta_(t)S^(@) = - 0.1 K.J. K^(-1)."mole"^(-1) What Is the value of equilibrium constant ?