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Question

The free energy of the formation of $$'NO'$$ is $$78\ kJ/  mol $$ at the temperature of an automobile engine $$(1000K)$$. What is the $$K_{C}$$ equilibrium constant for this reaction at $$1000K$$.

$$\cfrac { 1 }{ 2 } { N }_{ 2 }(g)+\cfrac { 1 }{ 2 } { O }_{ 2 }(g)\rightleftharpoons  NO(g)$$


A
8.4×105
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B
7.1×109
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C
4.2×1010
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D
1.7×1019
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Solution

The correct option is B $$8.4\times {10}^{-5}$$
 The free energy of formation of $$NO$$ is $$78kJ\ { mol }^{ -1 }$$ at the temperature of an automobile engine $$(1000K)$$. The equilibrium constant for this reaction at $$1000K$$ is $$8.4\times {10}^{-5}$$.

$$\cfrac { 1 }{ 2 } { N }_{ 2 }(g)+\cfrac { 1 }{ 2 } { O }_{ 2 }(g)\rightleftharpoons NO(g)$$
$$\Delta { G }^{ o }=-2.303RT\log { { K }_{ eq} } $$

$$78kJ\quad { mol }^{ -1 } \times 1000 \: J/kJ =-2.303 \times 8.314 \: J/mol/K \: \times 1000 \: K \: \times \log   { K }_{ eq}  $$
$$ \log   { K }_{ eq} =-4.07 $$

$$   { K }_{ eq} =8.4\times {10}^{-5} $$

Hence, the correct option is $$\text{A}$$

Chemistry

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