Medium
For the reaction ${N_2}_{(g)} + 2{O_2}_{(g)} \rightleftharpoons 2{NO_2}_{(g)}$,the equilibrium constant is $100$. Find the equilibrium constant for the following reactions:
$(1)$ $2{NO_2}_{(g)} \rightleftharpoons {N_2}_{(g)} + 2{O_2}_{(g)}$
$(2)$ ${NO_2}_{(g)} \rightleftharpoons \frac{1}{2}{N_2}_{(g)} + {O_2}_{(g)}$
$(1)$ $2{NO_2}_{(g)} \rightleftharpoons {N_2}_{(g)} + 2{O_2}_{(g)}$
$(2)$ ${NO_2}_{(g)} \rightleftharpoons \frac{1}{2}{N_2}_{(g)} + {O_2}_{(g)}$
(N/A) Given reaction: ${N_2}_{(g)} + 2{O_2}_{(g)} \rightleftharpoons 2{NO_2}_{(g)}$,$K = 100$.
$(1)$ The reaction $2{NO_2}_{(g)} \rightleftharpoons {N_2}_{(g)} + 2{O_2}_{(g)}$ is the reverse of the given reaction.
Therefore,$K_1 = \frac{1}{K} = \frac{1}{100} = 0.01$.
$(2)$ The reaction ${NO_2}_{(g)} \rightleftharpoons \frac{1}{2}{N_2}_{(g)} + {O_2}_{(g)}$ is half of the reverse reaction.
Therefore,$K_2 = (K_1)^{1/2} = (0.01)^{1/2} = 0.1$.
$(1)$ The reaction $2{NO_2}_{(g)} \rightleftharpoons {N_2}_{(g)} + 2{O_2}_{(g)}$ is the reverse of the given reaction.
Therefore,$K_1 = \frac{1}{K} = \frac{1}{100} = 0.01$.
$(2)$ The reaction ${NO_2}_{(g)} \rightleftharpoons \frac{1}{2}{N_2}_{(g)} + {O_2}_{(g)}$ is half of the reverse reaction.
Therefore,$K_2 = (K_1)^{1/2} = (0.01)^{1/2} = 0.1$.
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