Home
Class 11
CHEMISTRY
Calculate the equilibrium concentration ...

Calculate the equilibrium concentration ratio of C to A if equimolar ratio of A and B were allowed to come to equilibrium at 300K.
`A(g)+B(g) iff C(g) +D(g) , DeltaG^(@)=`-830 cal.

Text Solution

Verified by Experts

The correct Answer is:
2
Promotional Banner

Topper's Solved these Questions

  • CHEMICAL EQUILIBRIUM

    NARENDRA AWASTHI|Exercise Level 1 (Q.93 To Q.122)|1 Videos
  • CHEMICAL EQUILIBRIUM

    NARENDRA AWASTHI|Exercise Level 2|1 Videos
  • ATOMIC STUCTURE

    NARENDRA AWASTHI|Exercise level 2|1 Videos
  • DILUTE SOLUTION

    NARENDRA AWASTHI|Exercise Level 3 - Match The Column|1 Videos

Similar Questions

Explore conceptually related problems

K_(p)=0.04 atm at 899K for the equilibrium shown below. What is the equilibrium concentration of C_(2)H_(6) when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium ? C_(6)H_(6)(g)hArrC_(2)H_(4)(g)+H_(2)(g)

Calculate the mole ratio of 240g of calcium and 240g of Magnesium.

What is the equilibrium concentration of each of the substances in the equilirbrium when the initial concentration of I Cl was 0.78M ? 2ICl(g)harrI_(2)(g)+Cl_(2)(g) , K_(c)=0.14

Give two examples for each of zero order and first order reactions. Write the equations for the rate of reaction in terms of concentration changes of reactants and products for the following ractions. 1) A(g) +B(g) to C(g) +D(g) 2) A(g) to B(g)+C(g) 3) A(g) +B(g)to C(g)

For A+BhArrC , the equilibrium concentrations of A and B at a temperature are 15 "mol" L^(-1) . When volume is doubled the reaction has equilibrium concentration of A is 10 "mol" L^(-1) . Calculate a. K_(c) b concentration of C in original equilibrium.

A(g) + 3(g) hArr 4C(g) Initial concentration of A is equal to that of B. The equilibrium concentration of A and C are equal. K_(c) is equal to .

For the reaction A + B hArr C + D , the concentrations of A and B are equal . The equilibrium concentration of C is twice that of A . K_(C) of the reaction is

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) Consider the two reactions: XeF_(6(g))+H_(2)O_((g)) harr XeOF_(4(g))+2HF_((g)), K_(1)" "XeO_(4(g))+XeF_(6(g)) harr XeOF_(4(g))+XeO_(3)F_(2(g)), K_(2) Then the equilibrium constant for the following reaction will be XeO_(4(g))+2HF_((g)) harr XeO_(3)F_(2(g))+H_(2)O_((g))

In the equilibrium reaction, A(g) + 2B(g) + (g), the equilibrium constant, K_(c) is given by the expression