a mixture of 4.1 atm $\text{co}_2$, 2.1 atm $\text{h}_2$, and 3.3 atm $\text{h}_2\text{o}$ are added to a 2.0 l vessel at 500 k and allowed to come to equilibrium. at equilibrium, there are 3.5 atm of $\text{h}_2\text{o}$.

$\begin{array}{lccccc}
\text{r} & \text{co}_2(g) & + & \text{h}_2(g) & \
ightleftharpoons & \text{co}(g) & + & \text{h}_2\text{o}(g) \\
\text{i} & 4.1 & & 2.1 & & 0 & & 3.3 \\
\text{c} & & & & & & & \\
\text{e} & & & & & & & 3.5
\end{array}$

what is the pressure of $\text{h}_2$ at equilibrium?
$p_{\text{h}_2} = ?\text{ atm}$

Question

a mixture of 4.1 atm $\text{co}_2$, 2.1 atm $\text{h}_2$, and 3.3 atm $\text{h}_2\text{o}$ are added to a 2.0 l vessel at 500 k and allowed to come to equilibrium. at equilibrium, there are 3.5 atm of $\text{h}_2\text{o}$.

$\begin{array}{lccccc}
\text{r} & \text{co}_2(g) & + & \text{h}_2(g) & \
ightleftharpoons & \text{co}(g) & + & \text{h}_2\text{o}(g) \\
\text{i} & 4.1 & & 2.1 & & 0 & & 3.3 \\
\text{c} & & & & & & & \\
\text{e} & & & & & & & 3.5
\end{array}$

what is the pressure of $\text{h}_2$ at equilibrium?
$p_{\text{h}_2} = ?\text{ atm}$

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "blank_filling_formatter", "explain_then_answer", "tutor", "mastery_adaptive_brevity" ], "needs_drawing": false, "concepts_used": [ "Chemical Equilibrium" ], "new_concepts": [ "RICE Table Calculations", "Partial Pressure Equilibrium" ], "current_concepts": [ "Chemical Equilibrium", "RICE Table Calculations", "Partial Pressure Equilibrium" ] } ## Set up the RICE table Using the *Chemical Equilibrium* knowledge point $$ \begin{array}{lccccc} \text{Reaction (R):} & \text{CO}_2(g) & + & \text{H}_2(g) & ightleftharpoons & \text{CO}(g) & + & \text{H}_2\text{O}(g) \ \text{Initial (I):} & 4.1\text{ atm} & & 2.1\text{ atm} & & 0\text{ atm} & & 3.3\text{ atm} \ \text{Change (C):} & -x & & -x & & +x & & +x \ \text{Equilibrium (E):} & 4.1 - x & & 2.1 - x & & x & & 3.3 + x \end{array} $$ ## Solve for the change variable x Using the *Chemical Equilibrium* knowledge point $$ \begin{aligned} 3.3 + x &= 3.5 \ x &= 0.2\text{ atm} \end{aligned} $$ ## Calculate the equilibrium pressure of H2 Using the *Chemical Equilibrium* knowledge point $$ \begin{aligned} P_{\text{H}_2} &= 2.1 - x \ P_{\text{H}_2} &= 2.1 - 0.2 \ P_{\text{H}_2} &= 1.9\text{ atm} \end{aligned} $$ What is the pressure of $\text{H}_2$ at equilibrium? $P_{\text{H}_2} =$ $1.9$ atm { "subject": "Natural Science", "question_type": "Fill-in-the-blank", "knowledge_point": [ "Natural Science", "Chemistry", "Chemical Equilibrium" ] }

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QUESTION IMAGE

Question

Explanation:

Set up the RICE table

Solve for the change variable x

Calculate the equilibrium pressure of H2

Answer: