Consider the following two equilibrium reactions:
$i$. $2NH_{3(g)} \rightleftharpoons N_{2(g)} + 3H_{2(g)}$
$ii$. $2ND_{3(g)} \rightleftharpoons N_{2(g)} + 3D_{2(g)}$
What is the difference in their equilibrium constants $(K_c)$?

  • A
    The equilibrium constant for reaction $i$ is greater than that for reaction $ii$.
  • B
    The equilibrium constant for reaction $i$ is less than that for reaction $ii$.
  • C
    The equilibrium constants are equal.
  • D
    The equilibrium constants cannot be compared.

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Similar Questions

From the given data of equilibrium constants for the following reactions:
$(1) \ CO_{2(g)} + H_{2(g)} \rightleftharpoons CO_{(g)} + H_2O_{(g)} \ ; \ K_1$
$(2) \ CO_{(g)} + H_2O_{(g)} \rightleftharpoons CO_{2(g)} + H_{2(g)} \ ; \ K_2$
Wait,the provided question text has a typo in the reaction equations. Assuming the standard problem format where we relate equilibrium constants for reverse or combined reactions,if the target reaction is the same as reaction $(1)$,the answer is $K_1$. However,based on the options provided,this is likely a question asking for the relationship between $K_1$ and $K_2$ where reaction $(2)$ is the reverse of reaction $(1)$. If reaction $(2)$ is the reverse of reaction $(1)$,then $K_2 = \frac{1}{K_1}$. Given the options,please re-verify the input. Assuming the question asks for the equilibrium constant of a reaction derived from these,if the target reaction is $CO_{(g)} + H_2O_{(g)} \rightleftharpoons CO_{2(g)} + H_{2(g)}$,the answer is $K_1^{-1}$. Given the options,if we assume the target reaction is the reverse of reaction $(1)$,then $K = \frac{1}{K_1}$.

Ammonia under a pressure of $15 \ atm$ at $27 \ ^{\circ}C$ is heated to $347 \ ^{\circ}C$ in a closed vessel in the presence of a catalyst. Under these conditions,$NH_3$ is partially decomposed according to the equation,$2NH_3 \rightleftharpoons N_2 + 3H_2$. The vessel is such that the volume remains constant,and the pressure increases to $50 \ atm$. Calculate the percentage of $NH_3$ actually decomposed.

For the reaction $A_{(g)} \rightleftharpoons B_{(g)} + C_{(g)}$,$A$ is $33 \%$ dissociated at a total pressure $P$. The correct relation between $P$ and $K_{p}$ is

$2$ moles of $N_2$ are mixed with $6$ moles of $H_2$ in a closed vessel of $1 \ L$ capacity. If $50\%$ of $N_2$ is converted into $NH_3$ at equilibrium,find the value of $K_c$ for the reaction: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$

Assign $A, B, C, D$ from the given type of reaction.
$PbCl_2 \downarrow + H_2SO_4 \rightleftharpoons PbSO_4 \downarrow + 2HCl$

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