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The value of $$K_p$$ for the reaction, $$CO_2(g)+C(s)\rightleftharpoons 2CO(g)$$ is $$3.0$$ at $$1000$$K. If initially $$P_{CO_2}=0.48$$ bar and $$P_{CO}=0$$ bar and pure graphite is present, calculate the equilibrium partial pressure of CO and $$CO_2$$.


Solution

The given reaction is :-
                         $${ CO }_{ 2 }\left( g \right) +C\left( s \right) \rightleftharpoons 2CO\left( g \right) \quad \quad ;\quad { K }_{ P }=3$$
Initial pressure : $$0.48$$            $$0$$                 $$0$$
At eqm :       $$(0.48-p)$$                           $$2p$$                      (let)
Now,  $${ K }_{ P }=\dfrac { { \left[ CO \right]  }^{ 2 } }{ \left[ { CO }_{ 2 } \right]  } \quad \quad { K }_{ P }=\dfrac { { \left( pCO \right)  }^{ 2 } }{ \left( { pCO }_{ 2 } \right)  } $$
$$\Rightarrow \quad { K }_{ P }=\dfrac { { \left( 2p \right)  }^{ 2 } }{ \left( 0.48-p \right)  } $$
$$\Rightarrow \quad 3\left( 0.48-p \right) ={ 4p }^{ 2 }$$
$$\Rightarrow \quad { 4p }^{ 2 }+3p-1.44=0$$
$$\Rightarrow \quad p=\dfrac { -3\pm \sqrt { 9+23.04 }  }{ 8 } =\dfrac { -3\pm 5.66 }{ 8 } $$
$$\Rightarrow \quad p=\dfrac { 2.66 }{ 8 } =0.3325bar$$
So, Partial pressure of $$CO$$ at eqm $$=2\times 0.3325=0.665bar$$
      Partial pressure of $${ CO }_{ 2 }$$ at eqm $$=0.48-0.3325=0.1475bar$$.

Chemistry

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