In a closed container for the reaction $$H2(g)+I2(g)$$⇌$$2HI(g)$$ the equilibrium concentrations $$H2=0.5M$$,$$I2=0.5M$$ and [HI]$$=1.25M$$. If $$0.5$$ moles of HI was added to $$1$$ L of equilibrium mixture, then $$H2$$ at new equilibrium mixture is

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In a closed container for the reaction $$H2(g)+I2(g)$$⇌$$2HI(g)$$ the equilibrium concentrations $$H2=0.5M$$,$$I2=0.5M$$ and [HI]$$=1.25M$$. If $$0.5$$ moles of HI was added to $$1$$ L of equilibrium mixture, then $$H2$$ at new equilibrium mixture is

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Equilibrium constant $$KC=$$[HI]$$2H2I2$$ $$=$$[$$1.25$$]$$2$$[$$0.5$$][$$0.5$$] $$=6.25$$                                $$H22$$,$$+I2(g)$$⇌$$2HI$$  at old equilibrium     $$0.5$$     $$0.5$$     $$1.25$$ at new equilibrium $$+x20.5+x2+x20.5+x2$$  $$-x1.75-x$$ $$(1.75-x)20.5+x20.5+x2=6.25$$ $$1.75-x0.5+x2=2.5$$ $$1.75-x=1.25+1.25$$ x $$2.25$$ $$x=0.5$$ $$x=0.222$$ Equilibrium concentration of $$H2=0.61M$$

Chemical Equilibrium

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In a closed container for the reaction $H_{2_{(g)}} + I_{2_{(g)}} ⇌ 2 H I_{(g)}$ the equilibrium concentrations $[H_{2}] = 0.5 M, [I_{2}] = 0.5 M and [H I] = 1.25 M$ . If 0.5 moles of HI was added to 1 L of equilibrium mixture, then $[H_{2}]$ at new equilibrium mixture is

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Chemical Equilibrium

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In a closed container for the reaction H2(g)+I2(g)⇌2HI(g) the equilibrium concentrations H2=0.5M,I2=0.5M and [HI]=1.25M. If 0.5 moles of HI was added to 1 L of equilibrium mixture, then H2 at new equilibrium mixture is

Equilibrium constant KC=[HI]2H2I2 =[1.25]2[0.5][0.5] =6.25 H22,+I2(g)⇌2HI at old equilibrium 0.5 0.5 1.25 at new equilibrium +x20.5+x2+x20.5+x2 -x1.75-x (1.75-x)20.5+x20.5+x2=6.25 1.75-x0.5+x2=2.5 1.75-x=1.25+1.25 x 2.25 x=0.5 x=0.222 Equilibrium concentration of H2=0.61M

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In a closed container for the reaction $H_{2_{(g)}} + I_{2_{(g)}} ⇌ 2 H I_{(g)}$ the equilibrium concentrations $[H_{2}] = 0.5 M, [I_{2}] = 0.5 M and [H I] = 1.25 M$ . If 0.5 moles of HI was added to 1 L of equilibrium mixture, then $[H_{2}]$ at new equilibrium mixture is