For the reaction: CH4(g) + 2O2(g) --> 2H2O(l) + CO2(g) Will the rate of the reac

Question

For the reaction: CH4(g) + 2O2(g) --> 2H2O(l) + CO2(g) Will the rate of the reaction increase, decrease, or remain the same if the concentration of methane (CH4) increases? For the reaction: CH4(g) + 2O2(g) --> 2H2O(l) + CO2(g) Will the rate of the reaction increase, decrease, or remain the same if the concentration of methane (CH4) increases? CH4(g) + 2O2(g) --> 2H2O(l) + CO2(g) Will the rate of the reaction increase, decrease, or remain the same if the concentration of methane (CH4) increases?

Explanation / Answer

reactants moles (3) = products moles (3)

according lechatlier's principle no effect of pressure

rate=k[CH4][O2]

The rate of reaction will remain the same. It doesnt matter how much of one reactant you add, because it can only react with a set amount of the other reactant. The rest that you added will just be floating around as excess reactant.

rate of reaction depends only on temperature because this is an exothermic reaction hence decreasing temperature favors forward the reaction.

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