If a sample of $N a H {C O}_{3} (s)$ is brought to a temperature of $100^{o} C$ in a closed container total gas pressure at equilibrium is:

0.96 a t m

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0.23 a t m

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0.48 a t m

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0.46 a t m

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Solution

The correct option is A $0.96 a t m$
Given,

$K_{p} = 0.23$

$Δ H^{\circ} = 136 k J$

The reaction given is $2 N a H C O_{3} (s) ⇌ N a_{2} C O_{3} (s) + C O_{2} (g) + H_{2} O (g)$

Now we know that $K_{p}$ for the above given reaction can be given as $K_{p} = (P_{C O_{2}})^{1} \times (P_{H_{2} O})^{1}$ , since only $H_{2} O$ and $C O_{2}$ are present in gaseous states. Now substituting, the value of $K_{p}$ in the equation we get,

$K_{p} = (P_{C O_{2}})^{1} \times (P_{H_{2} O})^{1}$

$0.23 = (P_{C O_{2}})^{1} \times (P_{H_{2} O})^{1}$

Also, since moles of $H_{2} O$ and $C O_{2}$ are the same, the partial pressures will also be the same. Let that partial pressure be $P$ .

Substituting the values in the formula we get,

$P^{2} = 0.23$

Taking the square root on both sides,

$P = 0.479$

Total gas pressure will then be $= P_{C O_{2}} + P_{H_{2} O} = 2 P$

$\Rightarrow 2 \times 0.479$

$\Rightarrow 0.958 a t m \approx 0.96 a t m$

Hence, the correct answer will be option A

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If a sample of NaHCO3(s) is brought to a temperature of 100oC in a closed container total gas pressure at equilibrium is:

If a sample of $N a H {C O}_{3} (s)$ is brought to a temperature of $100^{o} C$ in a closed container total gas pressure at equilibrium is: