2SO\(_2\)(g) + O\(_2\)(g) ↔ 2SO\(_3\)(g)
ΔH = -189 kJ mol-1
The equilibrium constant for the reaction above is increased by?
decreasing the temperature of the system
decrease the pressure of the system
The addition of a catalyst to the system
Increasing the surface area of the vessel
Explanation
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Discussions (20)
equilibrium constant k will be increased if the product is higher than the reactant "K= product /reactant , this is an exothermic reaction in which increase in temperature favoured the backward reaction and decrease in temperature increase the product , also increase in pressure favoured the forward reaction and decrease favoured the backward reaction and reciprocal for the endothermic reaction so with this the answer is " B"
U are wrong bcos equilibrium constant is only affected by change in temperature,so d ans is A
Myschool u guys have to change the answer because equillibrium constant is affected only by temprature
Temperature changes affect the value of equilibrium constant, Other factors such as concentration, pressure and catalysis do not affect K(eq).
The correct answer should be A because equilibrium constant is only affected by temperature
They are also other factors that affects equilibrium which are:
1: concentration and pressure of the reactants and products
2: temperature
For an exothermic reaction ( ΔH is negative), K decreases with an increase in temperature. For an endothermic reaction ( ΔH is positive), K increases with an increase in temperature.
