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where k f is the rate constant for the forward reaction and k b is the rate constant for the backward reaction and the square brackets, […], denote concentration. If only A is present at the beginning, time t = 0 , with a concentration [A] 0 , the sum of the two concentrations, [A] t and [B] t , at time t , will be equal to [A] 0 .
Equality of forward and backward reaction rates, however, is a necessary condition for chemical equilibrium, though it is not sufficient to explain why equilibrium occurs. Despite the limitations of this derivation, the equilibrium constant for a reaction is indeed a constant, independent of the activities of the various species involved ...
A reversible reaction is a reaction in which the conversion of reactants to products and the conversion of products to reactants occur simultaneously. [1]+ + A and B can react to form C and D or, in the reverse reaction, C and D can react to form A and B.
Guldberg and Waage also recognized that chemical equilibrium is a dynamic process in which rates of reaction for the forward and backward reactions must be equal at chemical equilibrium. In order to derive the expression of the equilibrium constant appealing to kinetics, the expression of the rate equation must be used.
The forward and reverse reactions are competing with each other and differ in reaction rates. These rates depend on the concentration and therefore change with the time of the reaction: the reverse rate gradually increases and becomes equal to the rate of the forward reaction, establishing the so-called chemical equilibrium.
Informally, the Kolmogorov forward equation addresses the following problem. We have information about the state x of the system at time t (namely a probability distribution p t ( x ) {\displaystyle p_{t}(x)} ); we want to know the probability distribution of the state at a later time s > t {\displaystyle s>t} .
The order of differencing can be reversed for the time step (i.e., forward/backward followed by backward/forward). For nonlinear equations, this procedure provides the best results. For linear equations, the MacCormack scheme is equivalent to the Lax–Wendroff method. [4]
A catalyst increases the rate of a reaction without being consumed in the reaction. The use of a catalyst does not affect the position and composition of the equilibrium of a reaction, because both the forward and backward reactions are sped up by the same factor. For example, consider the Haber process for the synthesis of ammonia (NH 3):