Home
Class 11
CHEMISTRY
For the reaction at 300 K A((g)) harr V(...

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 ?

A

0

B

1

C

10

D

0.1

Text Solution

Verified by Experts

The correct Answer is:
B
`Delta G^(@)=DeltaH^(@)-TDeltaS^(@)=-RTlnk`
Promotional Banner

Similar Questions

Explore conceptually related problems

For the reaction at 298K: A_((g)) +B_((g)) hArr C_((g)) + D_((g)) Delta H^(@) + 29.8kcal and Delta S^(@) = 100cal K^(-1) . Find the value of equilibrium constant.

For the hypothetical reaction A_(2(g)) + B_(2(g)) hArr 2AB_((g)) Delta_(r ) G^(@) and Delta_(r)S^(@) are 20 kJ/mol and -20 JK^(-1) mol^(-1) respectively at 200K. If Delta_(r)C_(P) is 20 JK^(-1) mol^(-1) then Delta_(r ) H^(@) at 400K is

H_(2(g)) rarr 2H_((g)) , DeltaH = 400 KJ , then DeltaS_("system") at 2000 K is

The following equilibrium constants were determined at 1120k : 2CO_((g)) hArr C_((s)) + CO_(2(g)) , Kp_(I) = 10^(-14) atm^(-1) , CO_((g)0 + Cl_(2(g)) hArr COCl_(2(g)) , Kp_(2) = 6 xx 10^(-3) atm^(-1) What is the equilibrium constant Kc for the following reaction at 1120K: C_((s)) + CO_(2(g)) + 2Cl_(2(g)) hArr 2CO_(2) Cl_((g))

For the reaction, Ag_(2)O_((s)) hArr 2Ag_((s)) + (1)/(2) O_(2(g)) Delta H, Delta S and T are 40.63 kJ mol^(-1), 108.8 JK^(-1) mol^(-1) and 373K respectively. Free energy change Delta G of the reaction will be _____

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)

Passage: When all the coefficients in a balanced chemical equation are multiplied by a constant factor X the equilibrium constant (originally K) becomes K^J . Similarly, when balanced equations are added together, the equilibrium constant for the combined process is equal to the product of the equilibrium constants for each step. Equilibrium constant of the reversed reaction is numerically equal to the reciprocal of the equilibrium constant of the original equation. Unit of K_p = ("atm")^(Deltan) , Unit of K_c =("mol" L^(-1))^(Deltan) The equilibrium constants for the following reactions at 1400 K are given: 2H_2O_((g)) harr 2H_(2(g))+O_(2(g)) , K_1=2.1 xx 10^(-13) 2CO_(2(g)) harr 2CO_((g))+O_(2(g)) , K_2=1.4 xx 10^(-12) Then the equilibrium constant K for the reaction, H_(2(g))+CO_(2(g)) harr CO_((g)) + H_2O_((g)) is

The value of log_(10)^(k) for the reaction : A rarr B , If Delta H^(@) = - 55.07 kj "mole"^(-1) at 298K. Delta S^(@) = 10 jK^(-1) "mole"^(-1) at 298 K , R = 8.314 jK^(-1) "mole"^(-1) and 2.303 xx 8.314 xx 298 = 5705 is