1
COMEDK 2024 Evening Shift
MCQ (Single Correct Answer)
+1
-0

At $$700 \mathrm{~K}$$, the Equilibrium constant value for the formation of $$\mathrm{HI}$$ from $$\mathrm{H}_2$$ and $$\mathrm{I}_2$$ is 49.0 . 0.7 mole of $$\mathrm{HI}(\mathrm{g})$$ is present at equilibrium. What will be the concentrations of $$\mathrm{H}_2$$ and $$\mathrm{I}_2$$ gases if we initially started with $$\mathrm{HI}(\mathrm{g})$$ and allowed the reaction to reach equilibrium at the same temperature?

A
0.1195
B
0.3442
C
0.4692
D
0.521
2
COMEDK 2023 Morning Shift
MCQ (Single Correct Answer)
+1
-0

Consider the following equilibrium,

$$\begin{aligned} & 2 \mathrm{No}(g) \rightleftharpoons \mathrm{N}_2+\mathrm{O}_2 ; \mathrm{K}_{\mathrm{G}}=2.4 \times 10^{20} \\ & \mathrm{No}(\mathrm{g})+\frac{1}{2} \mathrm{Br}_2(\mathrm{~g}) \rightleftharpoons \mathrm{NoBr}(\mathrm{g}) ; \mathrm{K}_{\mathrm{C}_2}=1.4 \end{aligned}$$

Calculate $$K_C$$ for the reaction,

$$\frac{1}{2} \mathrm{~N}_2(g)+\frac{1}{2} \mathrm{O}_2(g)+\frac{1}{2} \mathrm{Br}_2(g) \rightleftharpoons \mathrm{NOBr}(g)$$

A
$$8.96 \times 10^{-11}$$
B
$$9.48 \times 10^{-9}$$
C
$$8.08 \times 10^{-12}$$
D
$$8.96 \times 10^{11}$$
3
COMEDK 2023 Evening Shift
MCQ (Single Correct Answer)
+1
-0

The equilibrium constants for the reactions $$a, b$$, and $$c$$ are as given:

a) $$\mathrm{N}_2+3 \mathrm{H}_2=2 \mathrm{NH}_3: \mathbf{K}_1$$

b) $$\mathrm{N}_2+\mathrm{O}_2=2 \mathrm{NO}: \mathrm{K}_2$$

c) $$2 \mathrm{H}_2+\mathrm{O}_2=2 \mathrm{H}_2 \mathrm{O}: \mathbf{K}_3$$

What would be the Equilibrium constant for the reaction:

$$4 \mathrm{NH}_3+5 \mathrm{O}_2=4 \mathrm{NO}+6 \mathrm{H}_2 \mathrm{O} ; \mathbf{K}_{\mathbf{x}}$$

A
$$ \mathbf{K}_{\mathbf{x}}=\mathrm{K}_2{ }^2 \mathrm{~K}_3{ }^3 / \mathrm{K}_1{ }^2 $$
B
$$ \mathbf{K}_{\mathbf{x}}=\mathrm{K}_1 / \mathrm{K}_2 \mathrm{~K}_3 $$
C
$$ \mathbf{K}_{\mathbf{x}}=1 / \mathbf{K}_1{ }^2+\mathbf{K}_2{ }^2+\mathrm{K}_3{ }^3 $$
D
$$ \mathbf{K}_{\mathbf{x}}=\mathrm{K}_1{ }^2 / \mathrm{K}_2 \mathrm{~K}_3{ }^3 $$
4
COMEDK 2022
MCQ (Single Correct Answer)
+1
-0

For the reaction, H$$_2$$(g) + I$$_2$$(g) $$\rightleftharpoons$$ 2HI(g) the position of equilibrium can be shifted to the right by

A
addition of Hi
B
addition of both I$$_2$$ and HI
C
increasing temperature
D
addition of I$$_2$$
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