Consider the following reaction at $$ 1000^{\circ} \mathrm{C} $$ and 1 atm : $$ \mathrm{FeO}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{Fe}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) ; \Delta \mathrm{G}^{0}=12761.2-7.03 \mathrm{~T} / \mathrm{mol} $$. The equilibrium partial pressure of hydrogen gas is ___ atm (round off to two decimal plac

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To find the equilibrium partial pressure of hydrogen gas, we need to calculate the equilibrium constant ($$ K_p $$) for the given reaction and then use it to find the partial pressure of hydrogen gas.

Given:
- Reaction: $$ \mathrm{FeO}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{Fe}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) $$
- $$ \Delta \mathrm{G}^{0}=12761.2-7.03 \mathrm{~T} / \mathrm{mol} $$
- Temperature: $$ 1000^{\circ} \mathrm{C} $$ (which is 1273 K)
- Pressure: 1 atm

First, let's calculate the equilibrium constant ($$ K_p $$) using the Gibbs free energy change ($$ \Delta G^{0} $$):

$$ K_p = e^{-\frac{\Delta G^{0}}{RT}} $$

where:
- $$ K_p $$ is the equilibrium constant
- $$ \Delta G^{0} $$ is the standard Gibbs free energy change
- $$ R $$ is the gas constant (approximately 8.314 J/mol*K)
- $$ T $$ is the temperature in Kelvin

Substitute the given values into the equation:

$$ K_p = e^{-\frac{12761.2-7.03 \times 1273}{8.314 \times 1273}} $$

Now, let's calculate the value of $$ K_p $$ and then use it to find the equilibrium partial pressure of hydrogen gas.
Calculate the value by following steps:
- step0: Calculate:
  $$e^{-\left(\frac{\left(12761.2-7.03\times 1273\right)}{\left(8.314\times 1273\right)}\right)}$$
- step1: Remove the parentheses:
  $$e^{-\left(\frac{12761.2-7.03\times 1273}{8.314\times 1273}\right)}$$
- step2: Multiply the numbers:
  $$e^{-\left(\frac{12761.2-8949.19}{8.314\times 1273}\right)}$$
- step3: Multiply the numbers:
  $$e^{-\left(\frac{12761.2-8949.19}{10583.722}\right)}$$
- step4: Subtract the numbers:
  $$e^{-\left(\frac{3812.01}{10583.722}\right)}$$
- step5: Divide the terms:
  $$e^{-\frac{1906005}{5291861}}$$
- step6: Express with a positive exponent:
  $$\frac{1}{e^{\frac{1906005}{5291861}}}$$
- step7: Simplify:
  $$\frac{1}{\sqrt[5291861]{e^{1906005}}}$$
- step8: Multiply by the Conjugate:
  $$\frac{\sqrt[5291861]{e^{3385856}}}{\sqrt[5291861]{e^{1906005}}\times \sqrt[5291861]{e^{3385856}}}$$
- step9: Multiply the numbers:
  $$\frac{\sqrt[5291861]{e^{3385856}}}{e}$$
The calculated value of $$ K_p $$ is approximately 0.697553.

Now, let's use the equilibrium constant ($$ K_p $$) to find the equilibrium partial pressure of hydrogen gas. The equilibrium constant expression for the given reaction is:

$$ K_p = \frac{P_{\mathrm{H}_2\mathrm{O}}}{P_{\mathrm{H}_2}} $$

where:
- $$ P_{\mathrm{H}_2\mathrm{O}} $$ is the equilibrium partial pressure of water vapor
- $$ P_{\mathrm{H}_2} $$ is the equilibrium partial pressure of hydrogen gas

Given that the reaction is at 1 atm, we can assume that the equilibrium partial pressure of water vapor is also 1 atm. Therefore, the equilibrium partial pressure of hydrogen gas can be calculated as:

$$ P_{\mathrm{H}_2} = \frac{1}{K_p} $$

Substitute the calculated value of $$ K_p $$ into the equation to find the equilibrium partial pressure of hydrogen gas.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{1}{0.697553}$$
- step1: Convert the expressions:
  $$\frac{1}{\frac{697553}{1000000}}$$
- step2: Multiply by the reciprocal:
  $$\frac{1000000}{697553}$$
The equilibrium partial pressure of hydrogen gas is approximately 1.433583 atm (rounded off to two decimal places).
The equilibrium partial pressure of hydrogen gas is 1.43 atm.

Consider the following reaction at and 1 atm :
.
The equilibrium partial pressure of hydrogen gas is ___ atm (round off to two decimal plac

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Consider the following reaction at and 1 atm : . The equilibrium partial pressure of hydrogen gas is ___ atm (round off to two decimal plac