A solution can be both dilute and concentrated saturated and dilute saturated an

Question

A solution can be both dilute and concentrated saturated and dilute saturated and unsaturated supersaturated and saturated Express the equilibrium constant for the following reaction: 16 CH_3Cl(g)+8Cl_2(g) equivalent 16 CH_2Cl_2(g)+8 H_2(g) k=[CH_2Cl_2][H_2]/[CH_3Cl][Cl_2] k=[CH_2CL_2]16[H_2]^8/[CH_3Cl]16[Cl_2]^8 k=[CH_3Cl][Cl_2]/[CH_2Cl_2][H_2] k=[CH_3Cl]1/2[Cl_2]/[CH_Cl_]/2[H_2] The following reaction is exothermic. Which change will shift the equilibrium to the left? 2SO_2(g)+O_2(g) equivalent 2SO_3(g) raising the temperature adding argon decrease volume all of the above none of the above

Explanation / Answer

6) Solution may be both diluted and concentrated

7) K =( [CH2Cl2]^16 * [H2]^8 ) / ( [CH3Cl] ^ 16 * [Cl2] ^8)

8) all the above

9) dn = number of gas mole of products - number of gas moles of reactants

dn = 2 - 3 = -1

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